Mechanical Properties of Solids – Guided Solutions

Q1 Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?

Brief Theory

According to Newton’s law of universal gravitation, every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. Unlike electric charge, mass has only one sign (positive), so gravitational fields cannot be cancelled or shielded.

Objects in orbit experience continuous free fall under gravity. This produces the sensation of weightlessness even though gravity is still acting. Gravitational effects such as tides depend on how the gravitational field changes with distance rather than on the total force alone.

Solution Map

• Understand difference between electric shielding and gravitational interaction.
• Recognize that orbiting bodies are in free fall.
• Compare gravitational force and tidal force.
• Use relation for tidal effect: proportional to \( \frac{M}{r^3} \).

Solution

(a) Gravitational shielding
Electrical shielding is possible because electric charges can rearrange themselves on the surface of a conductor so that the electric field inside becomes zero.

Gravity behaves differently. Mass cannot rearrange itself to cancel gravitational fields because there is no negative mass. Therefore, gravitational fields cannot be screened or shielded. Even if a body is placed inside a hollow sphere, it will still experience the gravitational influence of surrounding masses.

(b) Gravity detection inside an orbiting spacecraft
An astronaut in an orbiting spacecraft experiences weightlessness because both the astronaut and the spacecraft are falling toward the Earth with the same gravitational acceleration. This condition is known as free fall.

Even if the space station is very large, the astronaut inside it will still share the same orbital motion as the station. Since there is no relative acceleration between the astronaut and the station, gravity cannot be detected by ordinary mechanical means.

Therefore, increasing the size of the space station does not allow the astronaut to detect gravity.

(c) Why the Moon produces stronger tides than the Sun

Although the Sun exerts a larger gravitational force on the Earth due to its enormous mass, tidal effects depend on the difference in gravitational pull across the Earth.

The tidal force varies approximately as

\[ \text{Tidal force} \propto \frac{M}{r^{3}} \]

The Sun is extremely massive but also very far away from the Earth. The Moon, though much less massive, is much closer. Because the tidal force depends on the inverse cube of distance, the Moon produces a larger difference in gravitational pull between the near and far sides of the Earth.

Hence, the Moon’s tidal effect on the Earth is greater than that of the Sun despite the Sun exerting a larger total gravitational force.

Q2 Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula \(–GMm(1/r_2 – 1/r_1)\) is more/less accurate than the formula \(mg(r_2 – r_1)\) for the difference of potential energy between two points \(r_2\) and \(r_1\) distance away from the centre of the earth.

Brief Theory

The acceleration due to gravity near the Earth arises from Newton’s law of universal gravitation. For a body at distance \(r\) from Earth’s centre,

\[ g = \frac{GM}{r^2} \]

Thus the value of \(g\) depends on the distance from the Earth's centre. When the distance changes (either above the surface or below it), the gravitational acceleration also changes.

Solution Map

• Use gravitational formula \(g=\frac{GM}{r^2}\).
• Check how \(r\) changes with altitude.
• Apply uniform density argument for depth.
• Examine cancellation of mass in \(F=GMm/r^2\).
• Compare exact gravitational potential formula with approximate formula.

Solution

(a) Acceleration due to gravity with altitude

At height \(h\) above the Earth’s surface, the distance from the Earth's centre becomes \(R+h\). Hence

\[ g_h = \frac{GM}{(R+h)^2} \]

Since \(R+h > R\), the denominator increases and therefore the value of \(g\) decreases.

Correct choice: decreases

(b) Acceleration due to gravity with depth

If the Earth is assumed to be a sphere of uniform density, only the mass enclosed within the radius contributes to gravitational attraction.

The variation of \(g\) with depth \(d\) is

\[ g_d = g\left(1-\frac{d}{R}\right) \]

Thus \(g\) decreases linearly as we move deeper and becomes zero at the Earth's centre.

Correct choice: decreases

(c) Dependence of \(g\) on mass

From Newton's law of gravitation,

\[ F = \frac{GMm}{r^2} \]

The acceleration of the body is

\[ a = \frac{F}{m} = \frac{GM}{r^2} \]

Thus the mass \(m\) of the falling body cancels out. Therefore the acceleration due to gravity depends only on the mass of the Earth and distance from its centre.

Correct choice: independent of mass of the body

(d) Comparison of potential energy formulas

The exact expression for gravitational potential energy difference between two points is

\[ \Delta U = -GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \]

The simpler formula

\[ \Delta U = mg(r_2-r_1) \]

assumes that \(g\) is constant. This approximation is valid only for small heights compared to the Earth's radius.

Therefore the first expression is more accurate.

Correct choice: more accurate

Q3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?

Brief Theory

The relation between the orbital period of a planet and its orbital radius is given by Kepler’s Third Law of Planetary Motion. It states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis (or orbital radius for nearly circular orbits).

\[T^2 \propto r^3\]

Thus planets closer to the Sun have smaller orbital radii and therefore revolve faster.

Solution Map

• Planet completes orbit twice as fast → time period becomes \(T/2\).
• Apply Kepler’s third law \(T^2 \propto r^3\).
• Compare Earth’s orbit with the new planet’s orbit.
• Find ratio of orbital radii.

Solution

Let

• Earth’s orbital radius = \(R\)
• Earth’s time period = \(T\)
• Planet’s orbital radius = \(r\)
• Planet’s time period = \(T/2\)

From Kepler’s third law,

\[ T^2 \propto R^3 \]

\[ \left(\frac{T}{2}\right)^2 \propto r^3 \]

Dividing the second relation by the first,

\[ \frac{(T/2)^2}{T^2} = \frac{r^3}{R^3} \]

\[ \frac{1}{4} = \frac{r^3}{R^3} \]

Taking cube root on both sides,

\[ \frac{r}{R} = \left(\frac{1}{4}\right)^{1/3} \]

\[ \frac{r}{R} = \left(\frac{1}{2}\right)^{2/3} \]

Numerically,

\[ \left(\frac{1}{2}\right)^{2/3} \approx 0.63 \]

Therefore, the orbital radius of the planet is about 0.63 times the Earth’s orbital radius. This means the planet must orbit significantly closer to the Sun in order to complete one revolution in half the Earth's orbital period.

Final Result: The orbital size of the planet is \[ r = 0.63\,R \] i.e., about 63% of Earth’s orbital radius.

Q4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is \(\mathrm{4.22 \times 10^8}\) m. Show that the mass of Jupiter is about one-thousandth that of the Sun.

Brief Theory

A satellite orbiting a planet is held in orbit by gravitational attraction. For a circular orbit, the gravitational force provides the necessary centripetal force.

Equating gravitational force and centripetal force,

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

Using orbital velocity \(v=\frac{2\pi r}{T}\), we obtain a useful relation for the mass of the central body.

M = \frac{4\pi^2 r^3}{G T^2}

This relation allows us to determine the mass of a planet if the orbital radius and time period of its satellite are known.

Solution Map

• Use orbital motion relation for planetary mass.
• Convert orbital period from days to seconds.
• Substitute the values of \(r\), \(T\), and \(G\).
• Compare obtained mass with the mass of the Sun.

Solution

For the satellite Io,

• Orbital radius \(r = 4.22 \times 10^8 \, \text{m}\)
• Orbital period \(T = 1.769\) days

First convert the orbital period into seconds:

\[ T = 1.769 \times 24 \times 3600 \] \[ T \approx 1.53 \times 10^5 \, \text{s} \]

Using the orbital mass relation,

\[ M_J = \frac{4\pi^2 r^3}{G T^2} \]

Substituting the numerical values:

\[ M_J = \frac{4\pi^2 (4.22\times10^8)^3} {(6.67\times10^{-11})(1.53\times10^5)^2} \] \[ M_J \approx 1.9 \times 10^{27} \, \text{kg} \]

The mass of the Sun is approximately

\[ M_\odot \approx 2.0 \times 10^{30} \, \text{kg} \]

Therefore,

\[ \frac{M_J}{M_\odot} = \frac{1.9 \times 10^{27}} {2.0 \times 10^{30}} \] \[ \frac{M_J}{M_\odot} \approx 9.5 \times 10^{-4} \]

Thus the mass of Jupiter is approximately

\(M_J \approx 10^{-3} M_\odot\)

Hence the mass of Jupiter is about one-thousandth of the mass of the Sun.

Q5 Let us assume that our galaxy consists of \(\mathrm{2.5 \times 10^{11}}\) stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be \(\mathrm{10^5}\) ly.

Brief Theory

Stars in a galaxy move in orbits around the galactic centre due to the gravitational attraction of the total galactic mass. If we approximate the mass of the galaxy as concentrated near the centre, the motion of a star can be treated like a satellite orbiting a massive central body.

For such gravitational orbits, the time period of revolution is

T = 2\pi \sqrt{\frac{r^3}{GM}}

Thus the orbital period depends on the distance from the galactic centre and the total mass enclosed within that orbit.

Solution Map

• Calculate the total mass of the Milky Way using number of stars.
• Convert distance from light years to metres.
• Use gravitational orbital period relation.
• Convert the final result from seconds to years.

Solution

The galaxy contains

\[ 2.5 \times 10^{11} \text{ stars} \]

Each star has approximately one solar mass

\[ M_\odot = 2.0 \times 10^{30}\,\text{kg} \]

Therefore the total galactic mass is

\[ M = 2.5 \times 10^{11} \times 2.0 \times 10^{30} \] \[ M = 5.0 \times 10^{41}\,\text{kg} \]

The diameter of the Milky Way is \(10^5\) light years, so a star located halfway from the centre has radius

\[ r = 5.0 \times 10^4 \text{ light years} \]

Using \(1\) light year \(= 9.46 \times 10^{15}\,\text{m}\),

\[ r = 5.0 \times 10^4 \times 9.46 \times 10^{15} \] \[ r \approx 4.73 \times 10^{20}\,\text{m} \]

Using the orbital period relation,

\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

Substituting the values

\[ G = 6.67 \times 10^{-11} \] \[ r = 4.73 \times 10^{20}\,\text{m} \] \[ M = 5.0 \times 10^{41}\,\text{kg} \] \[ T = 2\pi \sqrt{\frac{(4.73 \times 10^{20})^3}{6.67\times10^{-11} \times 5.0\times10^{41}}} \] \[ T \approx 3.5 \times 10^{16}\,\text{s} \]

Converting seconds into years

\[ 1 \text{ year} = 3.15 \times 10^7 \text{ s} \] \[ T = \frac{3.5 \times 10^{16}}{3.15 \times 10^7} \] \[ T \approx 1.1 \times 10^9 \text{ years} \]

Thus a star located about 50,000 light years from the centre of the Milky Way takes roughly

\( \approx 1 \) billion years to complete one revolution around the galactic centre.

Q6 Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Brief Theory

When the reference of gravitational potential energy is taken at infinity, the potential energy of any object bound to Earth becomes negative. For a satellite moving in a circular orbit, gravitational attraction provides the centripetal force required for circular motion.

From Newton’s law of gravitation and circular motion,

\[ \frac{GMm}{r^2}=\frac{mv^2}{r} \]

This leads to the orbital velocity relation

\[ v^2=\frac{GM}{r} \]

Solution Map

• Use orbital velocity relation for circular motion.
• Calculate kinetic and potential energies.
• Find total mechanical energy of satellite.
• Compare escape energy for orbiting and stationary objects.

Solution

(a) Relation between total energy and kinetic energy

Using the orbital velocity relation,

\[ v^2=\frac{GM}{r} \]

The kinetic energy of the satellite is

\[ K=\frac{1}{2}mv^2 \] \[ K=\frac{GMm}{2r} \]

The gravitational potential energy at distance \(r\) is

U = -\frac{GMm}{r}

Hence the total mechanical energy is

\[ E = K + U \] \[ E = \frac{GMm}{2r} - \frac{GMm}{r} \] \[ E = -\frac{GMm}{2r} \]

Thus

\[ E = -K \]

Therefore the total energy of an orbiting satellite is equal to the negative of its kinetic energy.

Correct choice: kinetic energy

(b) Energy required for escape

An orbiting satellite already possesses kinetic energy due to its orbital motion. Its total mechanical energy is negative:

\[ E=-\frac{GMm}{2r} \]

To escape Earth's gravitational field, the total mechanical energy must become zero. Therefore only an additional energy equal to \(|E|\) needs to be supplied.

However, a stationary object at the same height has

• Kinetic energy = 0
• Potential energy \(U=-\frac{GMm}{r}\)

Thus its total energy is more negative, and a larger amount of energy must be supplied to make the total energy zero.

Hence, the energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than that required to project a stationary object from the same height.

Correct choice: less

Q7 Does the escape speed of a body from the earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?

Brief Theory

Escape speed is the minimum speed required for a body to move away from Earth and reach infinity with zero final velocity. The condition for escape is obtained using conservation of mechanical energy: the initial kinetic energy must equal the gravitational potential energy required to overcome Earth’s attraction.

Equating kinetic and gravitational potential energies gives the escape speed.

v_e = \sqrt{\frac{2GM}{R}}

This expression shows that escape velocity depends only on the mass and radius of the planet from which the body is launched.

Solution Map

• Apply energy conservation to obtain escape velocity.
• Check whether mass of the object appears in the final expression.
• Examine whether direction of projection affects escape energy.
• Check dependence on launching height or location.

Solution

(a) Dependence on mass of the body

During the derivation, the mass \(m\) of the body cancels out from both kinetic and potential energy terms. Therefore the escape speed does not depend on the mass of the body.

Escape speed is independent of the mass of the body.

(b) Dependence on location of projection

Escape velocity depends on the distance from the Earth's centre. If the body is launched from a different location where this distance changes, the escape speed will also change.

Escape speed depends on the location.

(c) Dependence on direction of projection

Escape speed depends only on the magnitude of velocity and not on its direction. Gravitational attraction is a central force directed toward the Earth's centre, so the required escape energy is the same regardless of the direction of launch.

Escape speed is independent of direction.

(d) Dependence on height of launch

If the body is launched from a height \(h\) above the Earth's surface, the distance from the Earth's centre becomes \(R+h\). The escape speed then becomes

\[ v_e = \sqrt{\frac{2GM}{R+h}} \]

Since \(R+h\) is larger than \(R\), the escape velocity decreases with increasing height.

Escape speed depends on height.

Final conclusion:

• Independent of mass of the body
• Independent of direction of projection
• Depends on location and height of launch

Q8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Brief Theory

A comet orbiting the Sun experiences a gravitational force directed toward the Sun. Since this force always acts along the line joining the comet and the Sun, it is a central force. Central forces produce no torque about the centre, which leads to conservation of angular momentum.

The gravitational potential energy of a body in the Sun’s field is

U = -\frac{GMm}{r}

In an elliptical orbit the distance \(r\) between the comet and the Sun changes continuously. Hence several physical quantities vary along the orbit while some remain conserved.

Solution Map

• Recognize that solar gravity is a central force.
• Use Kepler’s second law (equal areas in equal times).
• Apply conservation of angular momentum.
• Use conservation of mechanical energy.

Solution

(a) Linear speed

The linear speed of the comet is not constant. When the comet is near the Sun (perihelion), gravitational attraction is stronger and the comet moves faster. When it is far from the Sun (aphelion), the speed decreases.

Linear speed: Not constant

(b) Angular speed

Angular speed also changes during the motion. According to Kepler’s second law, the comet sweeps equal areas in equal time intervals. When the comet is close to the Sun it sweeps angles more rapidly, so the angular speed increases. When it is far away, angular speed decreases.

Angular speed: Not constant

(c) Angular momentum

Since gravitational force is a central force, the torque about the Sun is zero. Therefore angular momentum of the comet about the Sun remains conserved throughout the orbit.

Angular momentum: Constant

(d) Kinetic energy

Kinetic energy depends on speed. Because the comet moves faster near the Sun and slower farther away, its kinetic energy changes continuously along the orbit.

Kinetic energy: Not constant

(e) Potential energy

The gravitational potential energy depends on the distance \(r\) from the Sun. Since this distance changes along the elliptical orbit, the potential energy also varies. It becomes more negative when the comet is closer to the Sun.

Potential energy: Not constant

(f) Total mechanical energy

In the absence of non-conservative forces, the total mechanical energy \(E = K + U\) remains constant. Although kinetic and potential energies vary, their sum remains fixed for the entire orbit.

Total energy: Constant

Summary

• Linear speed → Not constant
• Angular speed → Not constant
• Angular momentum → Constant
• Kinetic energy → Not constant
• Potential energy → Not constant
• Total mechanical energy → Constant

Q9 Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.

Brief Theory

Astronauts orbiting Earth experience a condition of apparent weightlessness. This occurs because both the astronaut and the spacecraft are in continuous free fall around the Earth.

In the absence of normal gravity, body fluids that usually collect in the lower parts of the body redistribute toward the upper body. In addition, the vestibular system (inner ear balance mechanism) does not receive normal gravitational cues, which affects orientation.

Solution Map

• Understand the effect of microgravity on body fluids.
• Examine how fluid redistribution affects the face and feet.
• Consider effects on blood pressure and head circulation.
• Analyze how lack of gravity affects orientation.

Solution

(a) Swollen feet

On Earth, gravity causes blood and body fluids to accumulate in the legs and feet. In space, this downward pooling disappears because gravity is effectively absent. As a result, swelling of the feet is unlikely to occur.

Swollen feet → Not likely

(b) Swollen face

In microgravity, fluids shift from the lower body toward the upper body and head. This causes puffiness in the face, commonly known as “puffy face syndrome” observed in astronauts.

Swollen face → Likely

(c) Headache

The increased fluid flow toward the head can raise pressure in the head and sinuses. During the initial adaptation period, astronauts may experience headaches as the body adjusts to microgravity.

Headache → Likely

(d) Orientational problem

The human sense of balance depends strongly on gravity acting on the inner ear. In microgravity the vestibular system cannot provide reliable orientation signals. This often causes disorientation and difficulty distinguishing up from down.

Orientational problem → Likely

Final Answer

• Swollen feet → No
• Swollen face → Yes
• Headache → Yes
• Orientational problem → Yes

Q10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b,(iii)c, (iv) 0.

Brief Theory

Gravitational intensity at a point is defined as the gravitational force per unit mass experienced by a test particle placed at that point. For an extended body, the net gravitational field is obtained by the vector sum of the contributions from all mass elements of the body.

When a mass distribution has symmetry, many components of the gravitational field cancel. The direction of the resultant field can often be determined using symmetry arguments without detailed calculation.

Solution Map

• Consider gravitational field contributions from all mass elements of the hemisphere.
• Use symmetry about the central vertical axis.
• Examine cancellation of horizontal components.
• Add remaining vertical components.

Solution

Correct answer: (iii) c

Q11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Brief Theory

The gravitational intensity at a point inside or outside a mass distribution is obtained by the vector sum of the gravitational fields produced by all mass elements. For symmetric points such as the centre, several components cancel due to symmetry. However, at an arbitrary point the symmetry is incomplete, so the resultant field must be determined by considering the relative distribution of surrounding mass.

Solution Map

• Consider gravitational contributions from all mass elements of the hemisphere.
• Check which components cancel due to symmetry.
• Identify directions where cancellation does not occur.
• Determine the direction of the resultant field.

Solution

At the arbitrary point \(P\), the gravitational field is produced by all mass elements of the hemispherical shell.

Unlike the centre, point \(P\) is not symmetrically located with respect to the entire mass distribution. Therefore, the gravitational field components do not cancel completely.

Since the entire hemispherical shell lies below the flat surface, all mass elements produce vertical components of gravitational intensity directed downward. These components add together.

In addition, point \(P\) lies to the left of the axis of symmetry. Thus the mass distribution is effectively greater on the right side relative to point \(P\). As a result, the horizontal components of gravitational intensity do not cancel and the resultant horizontal component points toward the right.

Combining the horizontal and vertical components, the net gravitational intensity at point \(P\) is directed downward and toward the right.

Hence the direction of the gravitational intensity corresponds to arrow f.

Correct answer: (iii) f

Q12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun \(= 2\times10^{30}\ kg\), mass of the earth \(= 6\times10^{24}\ kg\). Neglect the effect of other planets etc. (orbital radius \(= 1.5 \times 10^{11}\ m\)).

Brief Theory

When a rocket moves along the line joining the Earth and the Sun, it experiences gravitational attraction from both bodies. At a certain point between them, the gravitational pulls become equal in magnitude and opposite in direction. At this location the net gravitational force becomes zero.

The gravitational force between two masses is given by Newton’s law of gravitation.

F = \frac{GMm}{r^2}

At the equilibrium point, the gravitational forces due to the Earth and the Sun must have equal magnitude.

Solution Map

• Let the Earth–Sun distance be \(R\).
• Assume the rocket is at distance \(x\) from Earth.
• Write gravitational forces due to Earth and Sun.
• Equate the forces to locate the balance point.
• Substitute numerical values.

Solution

Let

• Distance between Earth and Sun \(R = 1.5 \times 10^{11}\,m\)
• Distance of rocket from Earth \(= x\)
• Distance of rocket from Sun \(= R-x\)

Gravitational force due to Earth:

\[ F_E = \frac{G M_E m}{x^2} \]

Gravitational force due to Sun:

\[ F_S = \frac{G M_S m}{(R-x)^2} \]

At the point where net force becomes zero

\[ F_E = F_S \] \[ \frac{G M_E m}{x^2} = \frac{G M_S m}{(R-x)^2} \]

Canceling common factors \(G\) and \(m\)

\[ \frac{M_E}{x^2} = \frac{M_S}{(R-x)^2} \] Substituting \[ M_E = 6\times10^{24},\quad M_S = 2\times10^{30} \] \[ \frac{6\times10^{24}}{x^2} = \frac{2\times10^{30}}{(R-x)^2} \] Taking square root \[ \frac{R-x}{x} = \sqrt{\frac{M_S}{M_E}} \] \[ \frac{R-x}{x} = \sqrt{\frac{2\times10^{30}}{6\times10^{24}}} \] \[ \frac{R-x}{x} \approx 577 \] Therefore \[ R-x = 577x \] \[ R = 578x \] \[ x = \frac{R}{578} \] Substituting \(R = 1.5\times10^{11}\) \[ x \approx \frac{1.5\times10^{11}}{578} \] \[ x \approx 2.6\times10^{8}\,m \]

Hence the gravitational forces of the Earth and the Sun balance at a point approximately

\(2.6 \times 10^8\ m\) from the centre of the Earth along the Earth–Sun line.

Q13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is \(1.5 \times 10^8\) km.

Brief Theory

The mass of the Sun can be estimated by studying the motion of the Earth around it. The Earth moves in a nearly circular orbit due to the gravitational attraction of the Sun. This gravitational force provides the centripetal force required for the Earth’s orbital motion.

By equating the gravitational force between the Sun and Earth with the centripetal force required for orbital motion, the mass of the Sun can be determined.

M = \frac{4\pi^2 r^3}{G T^2}

This relation allows us to estimate the mass of the central body from the orbital radius and time period of the orbiting object.

Solution Map

• Consider Earth moving in circular orbit around the Sun.
• Equate gravitational force with centripetal force.
• Use orbital radius and time period of Earth.
• Substitute numerical values to calculate Sun’s mass.

Solution

Let

• Mass of Sun = \(M_\odot\)
• Mass of Earth = \(m\)
• Orbital radius \(r = 1.5 \times 10^8\) km \(= 1.5 \times 10^{11}\) m
• Time period of Earth \(T = 1\) year \(= 3.15 \times 10^7\) s

Gravitational force between Sun and Earth:

\[ F = \frac{G M_\odot m}{r^2} \]

Centripetal force required for Earth’s circular motion:

\[ F = \frac{m v^2}{r} \] Since orbital velocity \[ v = \frac{2\pi r}{T} \] Therefore \[ F = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \]

Equating gravitational and centripetal forces

\[ \frac{G M_\odot m}{r^2} = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \] Canceling \(m\) and simplifying \[ M_\odot = \frac{4\pi^2 r^3}{G T^2} \]

Substituting numerical values

\[ r = 1.5 \times 10^{11}\,m \] \[ T = 3.15 \times 10^{7}\,s \] \[ G = 6.67 \times 10^{-11} \] \[ M_\odot = \frac{4\pi^2 (1.5 \times 10^{11})^3} {6.67\times10^{-11}(3.15\times10^7)^2} \] \[ M_\odot \approx 2.0 \times 10^{30}\,kg \]

Thus, by analysing the orbital motion of the Earth, the mass of the Sun is estimated to be approximately

\(2 \times 10^{30}\,kg\).

Q14 A Saturn year is 29.5 times the Earth year. How far is Saturn from the Sun if the Earth is \(1.50 \times 10^8\) km away from the Sun ?

Brief Theory

The motion of planets around the Sun follows Kepler’s third law of planetary motion. This law states that the square of the orbital period of a planet is proportional to the cube of its mean distance from the Sun.

\frac{T^2}{r^3} = \text{constant}

Thus, for two planets orbiting the same central body (the Sun),

\[ \frac{T_S^2}{T_E^2} = \frac{R_S^3}{R_E^3} \] where \(T\) is the orbital period and \(R\) is the orbital radius.

Solution Map

• Use Kepler’s third law.
• Relate Saturn’s orbital period to Earth’s.
• Find ratio of orbital radii.
• Multiply by Earth’s orbital radius.

Solution

Let

• Earth’s orbital radius \(R_E = 1.50 \times 10^8\) km
• Earth’s time period \(T_E\)
• Saturn’s orbital radius \(R_S\)
• Saturn’s time period \(T_S = 29.5\,T_E\)

Using Kepler’s third law

\[ \frac{T_S^2}{T_E^2} = \frac{R_S^3}{R_E^3} \] Substituting \(T_S = 29.5\,T_E\) \[ \frac{(29.5T_E)^2}{T_E^2} = \frac{R_S^3}{R_E^3} \] \[ (29.5)^2 = \frac{R_S^3}{R_E^3} \] Taking cube root \[ \frac{R_S}{R_E} = (29.5)^{2/3} \]

Evaluating

\[ (29.5)^{2/3} \approx 9.6 \] Thus \[ R_S = 9.6\,R_E \] Substituting \(R_E = 1.50 \times 10^8\) km \[ R_S = 9.6 \times 1.50 \times 10^8 \] \[ R_S \approx 1.44 \times 10^9 \text{ km} \]

Therefore, the average distance of Saturn from the Sun is approximately

\(1.44 \times 10^9\) km.

Q15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?

Brief Theory

The gravitational force acting on a body at distance \(r\) from the centre of the Earth is given by Newton’s law of gravitation.

F = \frac{GMm}{r^2}

Thus the gravitational force varies inversely with the square of the distance from the Earth’s centre. When the height above the surface increases, the distance from the centre increases and the gravitational force decreases.

Solution Map

• Weight at Earth’s surface gives gravitational force \(F_0\).
• Find the new distance from Earth’s centre.
• Use inverse-square dependence of gravitational force.
• Compute ratio of forces.

Solution

The weight of the body on Earth’s surface is

\[ F_0 = 63\,\text{N} \]

At the Earth’s surface, the distance from the centre is

\[ r = R \]

The body is raised to height

\[ h = \frac{R}{2} \]

Hence the new distance from the Earth's centre is

\[ r = R + h \] \[ r = R + \frac{R}{2} = \frac{3R}{2} \]

Since gravitational force varies as \(1/r^2\),

\[ \frac{F}{F_0} = \frac{R^2}{\left(\frac{3R}{2}\right)^2} \] \[ \frac{F}{F_0} = \frac{R^2}{\frac{9R^2}{4}} \] \[ \frac{F}{F_0} = \frac{4}{9} \] Therefore \[ F = \frac{4}{9}F_0 \] Substituting \(F_0 = 63\,N\) \[ F = \frac{4}{9} \times 63 \] \[ F = 28\,N \]

Hence the gravitational force acting on the body at height \(R/2\) above Earth’s surface is

28 N.

Q16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?

Brief Theory

If the Earth is assumed to have uniform density, the gravitational acceleration inside the Earth decreases linearly with depth. This occurs because only the mass enclosed within the radius contributes to gravitational attraction.

g_d = g\left(1 - \frac{d}{R}\right)

Here \(d\) is the depth below the Earth's surface and \(R\) is the Earth's radius. At the centre (\(d=R\)), gravity becomes zero.

Solution Map

• Use relation for gravity at depth.
• Substitute depth \(d = R/2\).
• Determine the new value of gravitational acceleration.
• Compute the corresponding weight.

Solution

The weight of the body on the Earth's surface is

\[ W = mg = 250\,N \]

Halfway down to the centre means the depth

\[ d = \frac{R}{2} \] Using the relation for gravity at depth \[ g' = g\left(1 - \frac{d}{R}\right) \] Substituting \(d = R/2\) \[ g' = g\left(1 - \frac{1}{2}\right) \] \[ g' = \frac{g}{2} \]

The new weight of the body is

\[ W' = mg' \] \[ W' = m\left(\frac{g}{2}\right) \] \[ W' = \frac{1}{2} mg \] Substituting \(mg = 250\,N\) \[ W' = \frac{1}{2} \times 250 \] \[ W' = 125\,N \]

Hence the body would weigh

125 N

at a point halfway to the centre of the Earth.

Q17 A rocket is fired vertically with a speed of \(5\ km\ s^{-1}\) from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth \(= 6.0 \times 10^{24}\ kg\); mean radius of the earth \(= 6.4 \times 10^{6}\ m\); \(G = 6.67 \times 10^{-11}\ N\,m^2\,kg^{-2}\).

Brief Theory

As the rocket moves upward, its kinetic energy decreases while its gravitational potential energy increases. At the highest point, the velocity becomes zero. The maximum height reached can therefore be obtained using the conservation of mechanical energy.

The gravitational potential energy of a body at distance \(r\) from the Earth's centre is

U = -\frac{GMm}{r}

Thus the total mechanical energy remains constant during the motion.

Solution Map

• Write total mechanical energy at Earth’s surface.
• Write total energy at the highest point.
• Apply conservation of mechanical energy.
• Solve for the maximum distance from Earth’s centre.
• Find the height above Earth’s surface.

Solution

Initial speed of rocket

\[ v_0 = 5\,km\,s^{-1} = 5\times10^3\,m\,s^{-1} \]

Mass of Earth

\[ M = 6.0\times10^{24}\,kg \] Radius of Earth \[ R = 6.4\times10^6\,m \]

Let the maximum distance from Earth's centre be \(r\). At the highest point the velocity becomes zero.

Initial total energy at Earth's surface: \[ E_i = \frac{1}{2}mv_0^2 - \frac{GMm}{R} \] Final total energy at highest point: \[ E_f = -\frac{GMm}{r} \] Using conservation of energy \[ \frac{1}{2}mv_0^2 - \frac{GMm}{R} = -\frac{GMm}{r} \] Canceling \(m\) \[ \frac{1}{2}v_0^2 - \frac{GM}{R} = -\frac{GM}{r} \] Rearranging \[ \frac{1}{r} = \frac{1}{R} - \frac{v_0^2}{2GM} \] Substituting values \[ \frac{1}{r} = \frac{1}{6.4\times10^6} - \frac{(5\times10^3)^2}{2(6.67\times10^{-11})(6.0\times10^{24})} \] \[ \frac{1}{r} \approx 1.56\times10^{-7} - 3.12\times10^{-8} \] \[ \frac{1}{r} \approx 1.25\times10^{-7} \] Taking reciprocal \[ r \approx 8.0\times10^6\,m \]

Thus the maximum distance from Earth's centre is

\[ r \approx 8.0\times10^6\,m \] Height above Earth's surface \[ h = r - R \] \[ h = 8.0\times10^6 - 6.4\times10^6 \] \[ h = 1.6\times10^6\,m \]

Hence the rocket rises to a height of

\(1.6\times10^6\ m \approx 1600\ km\)

before returning to the Earth.

Q18 The escape speed of a projectile on the earth’s surface is \(11.2\ km\ s^{-1}\). A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Brief Theory

Escape velocity is the minimum speed required for a body to escape from Earth's gravitational field and reach infinity with zero speed.

The escape velocity is related to gravitational potential energy as

v_e = \sqrt{\frac{2GM}{R}}

If the initial speed is greater than the escape speed, the body will reach infinity with some residual speed. This can be found using conservation of mechanical energy.

Solution Map

• Use energy conservation between Earth’s surface and infinity.
• Express gravitational potential energy using escape velocity.
• Substitute the given initial velocity.
• Calculate the speed at infinity.

Solution

Escape speed from Earth

\[ v_e = 11.2\ km\ s^{-1} \] Initial speed of the body \[ v_0 = 3v_e \]

At Earth's surface the total mechanical energy is

\[ E = \frac{1}{2}mv_0^2 - \frac{GMm}{R} \]

Using the relation for escape velocity

\[ \frac{GMm}{R} = \frac{1}{2}mv_e^2 \] Substituting \[ E = \frac{1}{2}m(3v_e)^2 - \frac{1}{2}mv_e^2 \] \[ E = \frac{1}{2}m(9v_e^2 - v_e^2) \] \[ E = 4mv_e^2 \]

Far away from Earth the gravitational potential energy becomes zero. Hence the total energy becomes purely kinetic.

\[ E = \frac{1}{2}mv_\infty^2 \] Equating energies \[ \frac{1}{2}mv_\infty^2 = 4mv_e^2 \] Canceling \(m\) \[ v_\infty^2 = 8v_e^2 \] \[ v_\infty = \sqrt{8}v_e = 2\sqrt{2}v_e \] Substituting \(v_e = 11.2\ km\ s^{-1}\) \[ v_\infty = 2\sqrt{2} \times 11.2 \] \[ v_\infty \approx 31.7\ km\ s^{-1} \]

Hence the speed of the body far away from Earth is

\(31.7\ km\ s^{-1}\).

Q19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = \(6.0\times10^{24}\ kg\); radius of the earth = \(6.4 \times 10^6\ m\); \(G = 6.67 \times 10^{-11}\ N\,m^2\,kg^{-2}\).

Brief Theory

A satellite in circular orbit already possesses kinetic energy and gravitational potential energy. The total mechanical energy of a satellite in circular orbit is negative and is given by

E = -\frac{GMm}{2r}

To remove the satellite completely from Earth’s gravitational field, its total energy must become zero (energy at infinity). Therefore the additional energy required equals the magnitude of its orbital energy.

Solution Map

• Calculate orbital radius \(r = R + h\).
• Use orbital energy formula.
• Find magnitude of total orbital energy.
• This equals the energy required to escape.

Solution

Mass of satellite

\[ m = 200\,kg \] Mass of Earth \[ M = 6.0\times10^{24}\,kg \] Radius of Earth \[ R = 6.4\times10^6\,m \] Height of satellite \[ h = 400\,km = 4.0\times10^5\,m \] Orbital radius \[ r = R + h \] \[ r = 6.4\times10^6 + 4.0\times10^5 \] \[ r = 6.8\times10^6\,m \]

Total mechanical energy of the satellite in orbit

\[ E = -\frac{GMm}{2r} \]

Energy required to escape Earth’s gravitational field equals the magnitude of this energy:

\[ \Delta E = \frac{GMm}{2r} \] Substituting values \[ \Delta E = \frac{(6.67\times10^{-11})(6.0\times10^{24})(200)} {2(6.8\times10^6)} \] \[ \Delta E \approx 5.9\times10^9\,J \]

Thus the energy required to move the satellite from this orbit to infinity is

\(5.9 \times 10^9\ J\).

Q20 Two stars each of one solar mass (= \(2\times10^{30}\ kg\)) are approaching each other for a head-on collision. When they are a distance \(10^9\ km\), their speeds are negligible. What is the speed with which they collide ? The radius of each star is \(10^4\ km\). Assume the stars to remain undistorted until they collide.

Brief Theory

When two massive bodies approach each other under gravitational attraction, their gravitational potential energy decreases while their kinetic energy increases. The total mechanical energy of the system remains constant.

The gravitational potential energy between two masses separated by distance \(r\) is

U = -\frac{GM_1M_2}{r}

Thus the loss in potential energy during approach converts into kinetic energy of the two stars.

Solution Map

• Write initial separation between the stars.
• Determine final separation at collision.
• Use conservation of mechanical energy.
• Find the speed of each star just before collision.
• Compute the relative collision speed.

Solution

Mass of each star

\[ M = 2.0\times10^{30}\ kg \] Initial separation \[ r_i = 10^9\ km = 10^{12}\ m \] Radius of each star \[ R = 10^4\ km = 10^7\ m \]

At collision the distance between centres is

\[ r_f = 2R \] \[ r_f = 2\times10^7\ m \]

Initially the stars have negligible speed, so the kinetic energy is approximately zero.

Initial gravitational potential energy \[ U_i = -\frac{GM^2}{r_i} \] Final gravitational potential energy \[ U_f = -\frac{GM^2}{r_f} \]

If the speed of each star before collision is \(v\), the total kinetic energy of the system is

\[ K = 2 \times \frac{1}{2}Mv^2 = Mv^2 \] Applying conservation of energy \[ Mv^2 = GM^2\left(\frac{1}{r_f} - \frac{1}{r_i}\right) \] Cancel \(M\) \[ v^2 = GM\left(\frac{1}{r_f} - \frac{1}{r_i}\right) \] Substituting values \[ v^2 = (6.67\times10^{-11}) (2.0\times10^{30}) \left(\frac{1}{2\times10^7} - \frac{1}{10^{12}}\right) \] Since \[ \frac{1}{10^{12}} \ll \frac{1}{2\times10^7} \] \[ v^2 \approx 6.67\times10^{-11} \times 2.0\times10^{30} \times 5.0\times10^{-8} \] \[ v^2 \approx 6.7\times10^{12} \] \[ v \approx 2.6\times10^6\ m/s \]

This is the speed of each star just before collision.

Since the stars move toward each other with equal speeds, their relative collision speed is

\[ v_{\text{collision}} = 2v \] \[ v_{\text{collision}} \approx 5.2\times10^6\ m/s \]

Hence the stars collide with a speed of approximately

\(5.2\times10^6\ m/s\).

Q21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?

Brief Theory

The gravitational field produced by a spherical mass outside the sphere behaves as if the entire mass were concentrated at its centre. Therefore each sphere can be treated as a point mass when calculating gravitational field and potential at the midpoint.

The gravitational potential at a distance \(r\) from a mass \(M\) is

V = -\frac{GM}{r}

Gravitational potentials add algebraically, whereas gravitational forces add vectorially.

Solution Map

• Determine distance of midpoint from each sphere.
• Calculate gravitational forces due to each sphere.
• Use symmetry to find the resultant force.
• Compute total gravitational potential.
• Analyze stability of equilibrium.

Solution

Mass of each sphere

\[ M = 100\,kg \] Distance between centres \[ d = 1.0\,m \]

Distance of midpoint from each sphere

\[ r = \frac{d}{2} = 0.50\,m \]

Since this point lies outside the spheres (\(0.50 > 0.10\ m\)), each sphere behaves like a point mass located at its centre.

Gravitational Force at Midpoint

Force on a test mass \(m\) due to one sphere

\[ F = \frac{GMm}{r^2} \]

Both spheres produce equal forces in opposite directions because the midpoint is symmetrically located between them.

\[ F_{net} = 0 \]

Hence the gravitational force at the midpoint is

zero.

Gravitational Potential at Midpoint

Potential due to one sphere \[ V_1 = -\frac{GM}{0.50} \] Total potential due to both spheres \[ V = V_1 + V_2 \] \[ V = -\frac{2GM}{0.50} \] Substituting \(G = 6.67\times10^{-11}\) \[ V = -\frac{2(6.67\times10^{-11})(100)}{0.50} \] \[ V = -400G \] \[ V \approx -2.67\times10^{-8}\ J\,kg^{-1} \]

Nature of Equilibrium

At the midpoint the net force is zero, so a body placed there is in equilibrium.

However, if the body is displaced slightly toward one sphere, the gravitational attraction from that sphere becomes stronger than that from the other sphere.

Thus the body moves further away from the midpoint instead of returning.

Therefore the equilibrium is

unstable equilibrium.

Final Results:

• Gravitational force at midpoint = 0
• Gravitational potential = \(-2.67\times10^{-8}\ J\,kg^{-1}\)
• Equilibrium = unstable