Gravitation NCERT Solutions, Complete PDF Content and Important Questions

Answer: (a) No. Gravitational force cannot be shielded by any material. It is independent of the intervening medium between two bodies. (b) Yes. If the space station is very large, the gravitational acceleration $g$ will vary significantly across the different parts of the space station. This measurable difference in gravity (tidal force) will allow the astronaut to detect gravity. (c) The gravitational force depends inversely on the square of the distance ($1/r^2$), whereas the tidal effect (which is the difference in gravitational force across the Earth's diameter) depends inversely on the cube of the distance ($1/r^3$). Since the Moon is much closer to the Earth than the Sun, the $1/r^3$ factor makes the Moon's tidal effect much stronger, even though the Sun's absolute gravitational pull is greater.

Answer: (a) Decreases. ($g_h = g(1 - \frac{2h}{R})$) (b) Decreases. ($g_d = g(1 - \frac{d}{R})$) (c) Mass of the body. ($g = \frac{GM}{R^2}$, where $M$ is the mass of the Earth, not the object). (d) More accurate. The formula $mg(r_2 - r_1)$ assumes $g$ is constant, which is only valid for small heights near the surface. The exact formula $-GMm(1/r_2 - 1/r_1)$ accounts for the variation of $g$ with distance.

Answer: Given: Time period of the planet $T_p = \frac{1}{2} T_e$ (since it goes twice as fast). According to Kepler's Third Law: $$\frac{T_p^2}{T_e^2} = \frac{R_p^3}{R_e^3}$$ Substituting $T_p = 0.5 T_e$: $$\frac{(0.5 T_e)^2}{T_e^2} = \frac{R_p^3}{R_e^3}$$ $$0.25 = \left(\frac{R_p}{R_e}\right)^3$$ $$\frac{R_p}{R_e} = (0.25)^{1/3} \approx 0.63$$ Result: The orbital size of the planet would be approximately 0.63 times the orbital size of the Earth.

Answer: Let $M_j$ be the mass of Jupiter, and $M_s$ be the mass of the Sun. Orbital period of Io ($T_I$) = $1.769 \text{ days} = 1.769 \times 24 \times 3600 \text{ s}$. Orbital radius of Io ($R_I$) = $4.22 \times 10^8 \text{ m}$. Orbital period of Earth ($T_e$) = $365.25 \text{ days} = 365.25 \times 24 \times 3600 \text{ s}$. Orbital radius of Earth ($R_e$) = $1.496 \times 10^{11} \text{ m}$. Using Kepler's modified third law ($T^2 = \frac{4\pi^2 r^3}{GM}$): For Jupiter and Io: $M_j = \frac{4\pi^2 R_I^3}{G T_I^2}$ For Sun and Earth: $M_s = \frac{4\pi^2 R_e^3}{G T_e^2}$ Dividing the two equations: $$\frac{M_j}{M_s} = \frac{R_I^3}{R_e^3} \times \frac{T_e^2}{T_I^2}$$ $$\frac{M_j}{M_s} = \left(\frac{4.22 \times 10^8}{1.496 \times 10^{11}}\right)^3 \times \left(\frac{365.25}{1.769}\right)^2$$ $$\frac{M_j}{M_s} \approx (2.82 \times 10^{-3})^3 \times (206.47)^2$$ $$\frac{M_j}{M_s} \approx (22.4 \times 10^{-9}) \times (4.26 \times 10^4) \approx 9.5 \times 10^{-4} \approx \frac{1}{1000}$$ Hence Proved , the mass of Jupiter is about one-thousandth that of the Sun.

Answer: Given: Mass of the galaxy ($M$) = $2.5 \times 10^{11}$ solar masses = $2.5 \times 10^{11} \times (2 \times 10^{30} \text{ kg}) = 5.0 \times 10^{41} \text{ kg}$. Radius of orbit ($r$) = 50,000 light years = $50,000 \times 9.46 \times 10^{15} \text{ m} = 4.73 \times 10^{20} \text{ m}$. Using the formula for orbital time period: $$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$$ $$T = \sqrt{\frac{4 \times (3.14)^2 \times (4.73 \times 10^{20})^3}{6.67 \times 10^{-11} \times 5.0 \times 10^{41}}}$$ $$T \approx \sqrt{\frac{39.43 \times 105.8 \times 10^{60}}{33.35 \times 10^{30}}}$$ $$T \approx \sqrt{1.25 \times 10^{32}} \text{ seconds}$$ $$T \approx 1.12 \times 10^{16} \text{ seconds}$$ Converting to years ($1 \text{ year} = 3.15 \times 10^7 \text{ s}$): $$T \approx \frac{1.12 \times 10^{16}}{3.15 \times 10^7} \approx 3.55 \times 10^8 \text{ years}$$ Result: The star will take approximately 355 million years to complete one revolution.

Answer: (a) Kinetic energy. (Total Energy $E = -K$, and $E = \frac{U}{2}$). (b) Less. An orbiting satellite already possesses significant kinetic energy due to its orbital motion. Therefore, it requires less additional energy to escape the Earth's gravity compared to a completely stationary object at the same height, which has zero kinetic energy.

Answer: (a) No: Escape speed ($v_e = \sqrt{\frac{2GM}{R}}$) does not contain the mass of the body ($m$). (b) Yes: The Earth is an oblate spheroid (not a perfect sphere), so the radius ($R$) varies slightly from the equator to the poles, altering the escape speed. (c) No: Escape speed is a scalar concept dependent on energy, so the direction of projection does not matter (assuming no air resistance). (d) Yes: Escape speed depends on the distance from the center of the Earth. From a height $h$, $v_e = \sqrt{\frac{2GM}{R+h}}$, so it decreases with height.

Answer: (a) Linear speed: No. It moves faster when closer to the Sun and slower when farther away. (b) Angular speed: No. Since the radius changes and speed changes, angular speed ($\omega = v/r$) is not constant. (c) Angular momentum: Yes. The gravitational force acts as a central force, so net torque is zero. Hence, angular momentum is conserved. (d) Kinetic energy: No. Since linear speed changes, kinetic energy changes. (e) Potential energy: No. Since the distance from the Sun changes, potential energy changes. (f) Total energy: Yes. By the law of conservation of energy, the sum of kinetic and potential energy remains constant.

Answer: B) The poles. Explanation: The Earth is flattened at the poles, making the polar radius smaller than the equatorial radius. Since $g \propto 1/R^2$, $g$ is maximum at the poles.

Answer: C) $F$. Explanation: The gravitational force between two masses is independent of the medium separating them. The universal gravitational constant $G$ remains unchanged.

Answer: B) $\sqrt{3}v_e$ Explanation: By conservation of energy: $\frac{1}{2}m(2v_e)^2 - \frac{GMm}{R} = \frac{1}{2}mv_{final}^2 + 0$. Since $\frac{GMm}{R} = \frac{1}{2}mv_e^2$, solving gives $v_{final} = \sqrt{3}v_e$.

Answer: The satellite and the astronaut inside it are both in a state of free fall towards the Earth with the same acceleration (centripetal acceleration). Since they are falling at the same rate, the normal reaction force exerted by the floor of the satellite on the astronaut is zero, causing the feeling of weightlessness.

Answer: A geostationary satellite orbits the Earth in the equatorial plane with a time period of 24 hours, exactly matching the Earth's rotation. Thus, it appears stationary from the ground. Use: Telecommunications and live TV broadcasting.

Answer: 1) Its time period must be exactly 24 hours. 2) It must orbit in the equatorial plane. 3) Its direction of rotation must be the same as the Earth's (West to East). 4) It must be at a specific height (~35,800 km above the surface).

Answer: Inside the Earth ($r < R$), $g$ is directly proportional to $r$ (straight line passing through the origin). Outside the Earth ($r > R$), $g$ is inversely proportional to $r^2$ (a curve decreasing rapidly). The peak is exactly at the surface ($r = R$).

Answer: Let Earth be a sphere of radius $R$, mass $M$, and uniform density $\rho$. At the surface: $g = \frac{GM}{R^2} = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3}\pi G R \rho$. At a depth $d$, the distance from the center is $(R-d)$. Only the mass of the inner sphere of radius $(R-d)$ pulls the object. Mass of inner sphere $M' = \frac{4}{3}\pi (R-d)^3 \rho$. Gravity at depth $d$: $g_d = \frac{GM'}{(R-d)^2} = \frac{4}{3}\pi G (R-d) \rho$. Dividing the two equations: $\frac{g_d}{g} = \frac{R-d}{R} = 1 - \frac{d}{R}$. Therefore, $g_d = g(1 - \frac{d}{R})$. At the center of the Earth, $d = R$. Substituting this gives $g_d = g(1 - 1) = 0$. Since weight is $mg$, weight becomes zero.

Answer: Escape velocity is the minimum initial velocity that must be given to a body so it completely escapes the planet's gravitational field and never returns. Derivation: Let a body of mass $m$ be projected with velocity $v_e$. Kinetic energy at surface = $\frac{1}{2} m v_e^2$. Gravitational Potential Energy at surface = $-\frac{GMm}{R}$. Total Energy at surface = $\frac{1}{2} m v_e^2 - \frac{GMm}{R}$. At infinity, both potential and kinetic energy become zero. By conservation of energy: $\frac{1}{2} m v_e^2 - \frac{GMm}{R} = 0$ $\frac{1}{2} m v_e^2 = \frac{GMm}{R}$ $v_e = \sqrt{\frac{2GM}{R}}$ Since $g = \frac{GM}{R^2}$, we can also write $v_e = \sqrt{2gR}$.

Answer: Kepler's Second Law states that the areal velocity of a planet is constant. Let a planet of mass $m$ move in an elliptical orbit around the Sun. In a small time interval $\Delta t$, the planet moves a distance $\Delta \vec{r} = \vec{v} \Delta t$. The area $\Delta A$ swept by the radius vector $\vec{r}$ is a small triangle: $\Delta \vec{A} = \frac{1}{2} (\vec{r} \times \Delta \vec{r}) = \frac{1}{2} (\vec{r} \times \vec{v} \Delta t)$. Dividing by $\Delta t$: Areal velocity $\frac{\Delta \vec{A}}{\Delta t} = \frac{1}{2} (\vec{r} \times \vec{v})$. Multiply and divide by $m$: $\frac{\Delta \vec{A}}{\Delta t} = \frac{1}{2m} (\vec{r} \times m\vec{v}) = \frac{\vec{L}}{2m}$. Since the gravitational force acts along the line joining the planet and the Sun, torque ($\tau = \vec{r} \times \vec{F}$) is zero. Hence, angular momentum $\vec{L}$ is constant. Since $\vec{L}$ and $m$ are constant, the areal velocity $\frac{\vec{L}}{2m}$ is constant. Direction: Read the passage and answer questions 11-13. India's ISRO successfully launched a polar satellite to monitor weather patterns. Polar satellites orbit Earth at a lower altitude (around 500-800 km) compared to geostationary satellites. They pass over the north and south poles while the Earth rotates underneath them, allowing them to scan the entire globe over time.

Answer: Polar satellites orbit at a much lower altitude. According to Kepler's Third Law ($T^2 \propto R^3$), a smaller orbital radius results in a much shorter orbital time period.

Answer: No. Polar satellites continuously move relative to a point on the Earth's surface. For DTH broadcasting, antennas are fixed in one direction; hence, a satellite that appears stationary (Geostationary satellite) is required.

Answer: For a satellite very close to Earth, $v_o = \sqrt{gR}$. $v_o = \sqrt{9.8 \times 6.4 \times 10^6} = \sqrt{62.72 \times 10^6} \approx 7.92 \times 10^3 \text{ m/s} = \textbf{7.92 km/s}$. (Directions: Select A if both are true and Reason is correct explanation; Select B if both are true but Reason is not the correct explanation; Select C if Assertion is true but Reason is false; Select D if Assertion is false but Reason is true).

Answer: A . Both A and R are true, and R correctly explains the physical significance of negative total energy (a bound system).

Answer: D . The Assertion is false, but the Reason is true. Since gravity is weaker on the Moon (1/6th of Earth's), the ball will encounter less downward pull and will travel to a greater maximum height ($H = u^2/2g$).