Free NCERT Solutions, Notes & MCQs for Class 9–12

Q1: What is gravitation?

Gravitation is the universal force of attraction that acts between any two bodies having mass. It is responsible for the falling of objects on Earth and the motion of planets and satellites.

Q2: State Newton’s universal law of gravitation.

Every mass in the universe attracts every other mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Q3: What is the SI unit of gravitational constant (G)?

The SI unit of gravitational constant (G) is newton square metre per kilogram square \((\text{N m}^2 \text{kg}^{-2})\).

Q4: Define acceleration due to gravity.

Acceleration due to gravity is the acceleration produced in a body when it falls freely under the gravitational pull of the Earth.

Q5: What is the average value of (g) on Earth?

The average value of acceleration due to gravity on the Earth’s surface is approximately \(9.8,\text{m s}^{-2}\).

Q6: What is gravitational potential energy?

Gravitational potential energy is the energy possessed by a body due to its position in a gravitational field.

Q7: What is escape speed?

Escape speed is the minimum speed required by a body to escape completely from the gravitational field of a planet without further propulsion.

Q8: What is meant by a satellite?

A satellite is a body that revolves around a planet under the influence of gravitational force.

Q9: Why is gravitational force always attractive?

Gravitational force arises due to mass, and mass only attracts mass; therefore, gravity is always attractive in nature.

Q10: What happens to (g) at the centre of the Earth?

At the centre of the Earth, acceleration due to gravity becomes zero because gravitational forces from all directions cancel each other.

Q11: What is the shape of planetary orbits according to Kepler’s first law?

According to Kepler’s first law, planetary orbits are elliptical with the Sun at one focus.

Q12: What provides centripetal force to a satellite?

The gravitational force between the Earth and the satellite provides the required centripetal force.

Q13: Define gravitational field intensity.

Gravitational field intensity is the gravitational force experienced per unit mass at a point in the field.

Q14: Why is weight different from mass?

Mass is constant everywhere, whereas weight depends on the local value of acceleration due to gravity.

Q15: What is the nature of gravitational potential energy at infinity?

Gravitational potential energy at infinity is taken as zero by convention.

Q1: Write the mathematical expression for Newton’s law of gravitation and explain each term.

The gravitational force between two masses is given by \(F = G \frac{m_1 m_2}{r^2}\). Here, \(F\) is the force of attraction, \(m_1\) and \(m_2\) are the interacting masses, \(r\) is the distance between their centres, and \(G\) is the universal gravitational constant.

Q2: Why is gravitational constant called universal?

Gravitational constant is called universal because its value remains the same everywhere in the universe and does not depend on the nature of masses or the medium between them.

Q3: State two factors affecting acceleration due to gravity.

Acceleration due to gravity depends on the mass of the Earth and the distance of the body from the Earth’s centre.

Q4: Why does a satellite not fall on Earth?

A satellite does not fall on Earth because its tangential velocity causes it to continuously fall around the Earth instead of falling straight down.

Q5: Write the expression for escape speed in terms of (g) and (R).

The escape speed is given by \(v_e = \sqrt{2gR}\), where (g) is the acceleration due to gravity and \(R\) is Earth’s radius.

Q6: What is weightlessness in a satellite?

Weightlessness in a satellite occurs because both the satellite and the objects inside it fall freely under gravity with the same acceleration.

Q7: Why does (g) decrease with altitude?

As altitude increases, the distance from the Earth’s centre increases, reducing gravitational attraction and hence decreasing (g).

Q8: State Kepler’s second law.

Kepler’s second law states that the line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

Q9: Write the relation between orbital velocity and radius.

Orbital velocity of a satellite varies inversely with the square root of the radius of its orbit.

Q10: Why is gravitational potential energy negative?

Gravitational potential energy is negative because work is done by gravity when a body moves closer to the Earth from infinity.

Q1: Derive the expression for acceleration due to gravity at the Earth’s surface.

Using Newton’s law of gravitation, the force on a mass \(m\) at Earth’s surface is \(F = G\frac{Mm}{R^2}\). This force produces acceleration \(g\), so \(mg = G\frac{Mm}{R^2}\). Cancelling \(m\), we get \(g = G\frac{M}{R^2}\).

Q2: Explain variation of (g) with height.

At height \(h\), the distance from Earth’s centre becomes \(R+h\). Substituting in the gravitational formula shows that \(g\) decreases as height increases due to increased distance.

Q3: Explain why all bodies fall with same acceleration.

Since acceleration due to gravity depends only on Earth’s mass and radius, all bodies fall with the same acceleration regardless of their mass.

Q4: Obtain expression for gravitational potential energy near Earth’s surface.

Near Earth’s surface, the gravitational potential energy change is given by \( \Delta U = mgh \), derived from the general expression using an approximation.

Q5: State and explain Kepler’s third law.

Kepler’s third law states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit.

Q1: Explain Cavendish experiment and its significance.

\(\textbf{Cavendish Experiment and Its Significance}\)

The Cavendish experiment is a classic experiment performed to measure the universal gravitational constant \(G\). It demonstrated for the first time that gravitational attraction exists not only between large celestial bodies but also between small laboratory-sized masses.

\(\textbf{Principle of the Experiment}\)

The experiment is based on the gravitational attraction between two small masses and two large masses. This attraction produces a small torque in a suspended system. The angular twist produced in the suspension wire is used to calculate the gravitational force and hence the value of \(G\).

\(\textbf{Experimental Arrangement}\)

The main apparatus consists of a light horizontal rod suspended at its centre by a thin torsion wire. Two small lead spheres, each of mass \(m\), are fixed at the ends of the rod. Two large lead spheres, each of mass \(M\), are placed close to the small spheres.

When the large masses are brought near the small ones, gravitational attraction causes the rod to rotate slightly, twisting the torsion wire. A small mirror attached to the wire reflects a beam of light onto a distant scale, allowing the tiny angular deflection to be measured accurately.

\(\textbf{Theory and Working}\)

Let the distance between the centres of a small mass and a large mass be \(r\). The gravitational force between them is

\[ F = G\frac{Mm}{r^2} \]

This force produces a torque about the suspension axis. If \(l\) is the perpendicular distance of the force from the axis, the total gravitational torque acting on the system is

\[ \tau_g = 2Fl \]

The torsion wire provides a restoring torque proportional to the angle of twist \(\theta\),

\[ \tau_r = C\theta \] where \(C\) is the torsional constant of the wire. At equilibrium, \[ C\theta = 2Fl \] Substituting the value of \(F\), \[ C\theta = 2G\frac{Mm}{r^2}l \] Rearranging, \[ G = \frac{C\theta r^2}{2Mml} \]

Thus, by measuring \(\theta\), \(r\), \(M\), \(m\), and \(l\), the value of the gravitational constant \(G\) can be determined.

\(\textbf{Significance of the Cavendish Experiment}\)

The Cavendish experiment provided the first accurate value of the universal gravitational constant. Once \(G\) was known, it became possible to calculate the mass and average density of the Earth, which is why the experiment is often described as “weighing the Earth.”

The experiment proved that gravitational force is truly universal and acts between all masses, irrespective of their size. It also demonstrated the extraordinary sensitivity required to measure gravitational effects in the laboratory.

\(\textbf{Conclusion}\)

The Cavendish experiment occupies a central place in gravitational physics. It established the experimental foundation of Newton’s law of gravitation and linked terrestrial experiments with astronomical phenomena, making it one of the most significant experiments in classical physics.

Q2: Derive the expression for escape speed of Earth.

\(\textbf{Derivation of the Escape Speed of the Earth}\) Escape speed is the minimum speed that must be given to a body at the surface of the Earth so that it can escape completely from the Earth’s gravitational field and reach infinity with zero residual speed, without any further propulsion.

\(\textbf{Step 1: Energy of the body at the Earth’s surface}\) Consider a body of mass \(m\) projected vertically upward from the Earth’s surface with escape speed \(v_e\). Let the mass of the Earth be \(M\) and its radius be \(R\). The kinetic energy of the body at the Earth’s surface is \[ K = \frac{1}{2} m v_e^2 \] The gravitational potential energy at the Earth’s surface is \[ U = -\frac{GMm}{R} \] Hence, the total mechanical energy at the Earth’s surface is \[ E_{\text{surface}} = \frac{1}{2} m v_e^2 - \frac{GMm}{R} \] \(\textbf{Step 2: Energy of the body at infinity}\) At infinity, the gravitational potential energy is taken as zero and the body just comes to rest. Therefore, \[ E_{\infty} = 0 \] \(\textbf{Step 3: Application of conservation of mechanical energy}\) According to the law of conservation of energy, \[ E_{\text{surface}} = E_{\infty} \] \[ \frac{1}{2} m v_e^2 - \frac{GMm}{R} = 0 \] \(\)\textbf{Step 4: Expression for escape speed} Rearranging the above equation, \[ \frac{1}{2} m v_e^2 = \frac{GMm}{R} \] Cancelling \(m\) from both sides, \[ v_e^2 = \frac{2GM}{R} \] \[ \boxed{v_e = \sqrt{\frac{2GM}{R}}} \] \(\textbf{Relation with acceleration due to gravity}\) Since \[ g = \frac{GM}{R^2} \] we have \[ GM = gR^2 \] Substituting in the expression for escape speed, \[ v_e = \sqrt{2gR} \] \(\textbf{Numerical value for the Earth}\) For the Earth, \[ g = 9.8\,\text{m s}^{-2}, \quad R = 6.4 \times 10^6\,\text{m} \] \[ v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \] \[ \boxed{v_e \approx 11.2\,\text{km s}^{-1}} \] \bigskip \(\textbf{Conclusion}\) The escape speed of the Earth is independent of the mass of the body and depends only on the mass and radius of the Earth.

Q3: Explain motion and energy of an Earth satellite.

\(\textbf{Motion and Energy of an Earth Satellite}\) An Earth satellite is a body that revolves around the Earth under the influence of the Earth’s gravitational attraction. The motion of a satellite provides a clear application of Newton’s law of gravitation and the concept of circular motion. \(\textbf{Motion of an Earth Satellite}\) Consider a satellite of mass \(m\) revolving in a circular orbit of radius \(r\) around the Earth. Let the mass of the Earth be \(M\). The gravitational force between the Earth and the satellite is \[ F = G\frac{Mm}{r^2} \] This gravitational force acts towards the centre of the Earth and provides the necessary centripetal force required to keep the satellite in circular motion. Hence, \[ G\frac{Mm}{r^2} = \frac{mv^2}{r} \] Cancelling \(m\) and simplifying, \[ v = \sqrt{\frac{GM}{r}} \] This expression gives the orbital speed of the satellite. Thus, a satellite remains in orbit because its forward tangential velocity continuously balances the inward pull of gravity, causing it to fall around the Earth rather than straight towards it. \(\textbf{Energy of an Earth Satellite}\) The satellite possesses both kinetic energy and gravitational potential energy while in orbit. \(\textbf{Kinetic Energy}\) \[ K = \frac{1}{2}mv^2 \] Substituting the value of \(v\), \[ K = \frac{1}{2}m\left(\frac{GM}{r}\right) \] \[ K = \frac{GMm}{2r} \] \(\textbf{Gravitational Potential Energy}\) \[ U = -\frac{GMm}{r} \] The negative sign indicates that the satellite is bound to the Earth by gravitational attraction. \(\textbf{Total Mechanical Energy of the Satellite}\) The total energy \(E\) of the satellite is the sum of its kinetic and potential energies: \[ E = K + U \] \[ E = \frac{GMm}{2r} - \frac{GMm}{r} \] \[ E = -\frac{GMm}{2r} \] The total energy of an Earth satellite is therefore negative, which shows that the satellite is in a bound orbit and cannot escape the Earth’s gravitational field on its own. \(\textbf{Important Conclusions}\) \[ \begin{aligned} &1.\ \text{The gravitational force acts as the centripetal force for the satellite.} \\ &2.\ \text{The orbital speed and time period depend only on the orbital radius, not on the mass of the satellite.} \\ &3.\ \text{The total energy of a satellite in circular orbit is always negative.} \\ &4.\ \text{Satellites experience weightlessness because they are in continuous free fall.} \end{aligned} \]

Q4: Describe variation of acceleration due to gravity below the Earth’s surface.

\(\textbf{Variation of Acceleration Due to Gravity Below the Earth’s Surface}\)

The acceleration due to gravity does not remain constant inside the Earth. As we move below the Earth’s surface, the value of \(g\) decreases and finally becomes zero at the centre of the Earth.

\(\textbf{Assumptions}\)
The Earth is assumed to be a uniform sphere of radius \(R\) and mass \(M\). A body of mass \(m\) is located at a depth \(d\) below the Earth’s surface.

\(\textbf{Derivation}\)
At a depth \(d\), the distance of the body from the centre of the Earth is \[ r = R - d \] Only the mass of the Earth enclosed within radius \(r\) contributes to the gravitational force. Since mass is proportional to volume, the enclosed mass \(M_r\) is \[ M_r = M\left(\frac{r}{R}\right)^3 \] The gravitational force acting on the body at depth \(d\) is \[ F = G\frac{M_r m}{r^2} \] Substituting the value of \(M_r\), \[ F = G\frac{M m}{R^3} r \] The acceleration due to gravity at depth \(d\), denoted by \(g_d\), is \[ g_d = \frac{F}{m} \] \[ g_d = G\frac{M}{R^3} r \] Substituting \(r = R - d\), \[ g_d = G\frac{M}{R^3}(R - d) \] Since acceleration due to gravity at the surface is \[ g = G\frac{M}{R^2} \] we get \[ g_d = g\left(1 - \frac{d}{R}\right) \] \(\textbf{Physical Interpretation}\) This expression shows that acceleration due to gravity decreases linearly with depth below the Earth’s surface. When \(d = 0\), \(g_d = g\), which corresponds to the Earth’s surface. When \(d = R\), that is, at the centre of the Earth, \(g_d = 0\). \(\textbf{Conclusion}\) The acceleration due to gravity decreases uniformly as we move towards the centre of the Earth and becomes zero at the centre because the gravitational pulls from all directions cancel each other.

Q5: Discuss gravitational potential energy and its physical significance.

\(\textbf{Gravitational Potential Energy and Its Physical Significance}\)

Gravitational potential energy is the energy possessed by a body due to its position in a gravitational field. It arises because work must be done against the gravitational force to change the position of a body in the field of another massive body such as the Earth.

\(\textbf{Definition}\)

The gravitational potential energy of a body at a point is defined as the work done against the gravitational force in bringing the body slowly from infinity to that point, without changing its kinetic energy. By convention, the gravitational potential energy at infinity is taken to be zero.

\(\textbf{Expression for Gravitational Potential Energy}\) Consider a body of mass \(m\) at a distance \(r\) from the centre of the Earth of mass \(M\). The gravitational force acting on the body is \[ F = G\frac{Mm}{r^2} \] To move the body slowly outward through a small distance \(dr\), work done against gravity is \[ dW = F\,dr = G\frac{Mm}{r^2}dr \] The total work done in bringing the body from infinity to distance \(r\) is \[ U = -\int\limits_{\infty}^{r} G\frac{Mm}{r^2}dr \] Evaluating the integral, \[ U = -GMm \int\limits_{\infty}^{r} \frac{dr}{r^2} \] \[ U = -GMm\left[-\frac{1}{r}\right]_{\infty}^{r} \] \[ U = -\frac{GMm}{r} \] Thus, the gravitational potential energy of a body at distance \(r\) from the Earth’s centre is \[ U = -\frac{GMm}{r} \] \(\textbf{Gravitational Potential Energy Near the Earth’s Surface}\) For a body at a small height \(h\) above the Earth’s surface, where \(h \ll R\), \[ r = R + h \] \[ U = -\frac{GMm}{R+h} \] Using approximation, \[ U \approx -\frac{GMm}{R} + \frac{GMm}{R^2}h \] Since \(g = \frac{GM}{R^2}\), \[ U \approx -\frac{GMm}{R} + mgh \] Hence, the change in gravitational potential energy near the Earth’s surface is \[ \Delta U = mgh \] \(\textbf{Physical Significance of Gravitational Potential Energy}\)

The negative sign of gravitational potential energy indicates that the gravitational force is attractive and that the body is bound to the Earth. Energy must be supplied to the body to remove it completely from the Earth’s gravitational field.

Gravitational potential energy depends only on the position of the body and not on the path followed. This confirms that gravitational force is a conservative force and allows the use of energy conservation principles in gravitational problems.

The concept of gravitational potential energy plays a crucial role in understanding free fall, escape speed, satellite motion, and planetary orbits. It explains how kinetic and potential energies interchange during motion under gravity.

\(\textbf{Conclusion}\)
Gravitational potential energy provides a powerful energy-based description of motion in a gravitational field. It connects force, position, and energy, and forms the foundation for analyzing both terrestrial and celestial phenomena using the principle of conservation of mechanical energy.

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