GravitationClass 11 Physics NCERT Solutions

(a) No, a body cannot be shielded from the gravitational influence of nearby matter. Gravitational force is independent of the medium between two bodies and acts between any two masses in the universe. Unlike electrical forces, which can be shielded because of the existence of opposite charges (positive and negative) that can be rearranged to cancel the external field, there is no 'negative mass' or anti-gravity to counteract the gravitational force. Therefore, gravitational shielding is not possible.

(b) Yes, if the space station is very large, an astronaut can detect gravity. The sensation of weightlessness arises because both the astronaut and the spaceship are in a state of free fall towards the Earth with the same acceleration. However, in a very large space station, the acceleration due to gravity will not be uniform across its entire size. The part of the station closer to the Earth will experience a slightly stronger gravitational pull than the part farther away. This difference in gravitational force, known as a tidal force or gravity gradient, could be measured by sensitive instruments, allowing the astronaut to detect the presence of gravity.

(c) The tidal effect is caused by the difference in gravitational force exerted by a celestial body across the diameter of the Earth, not by the absolute strength of the force. The gravitational force is proportional to $1/r^2$, where $r$ is the distance to the celestial body. The tidal force, which is a differential force, is approximately proportional to $1/r^3$.

The Sun's gravitational pull on the Earth is indeed much greater than the Moon's. However, the Sun is much farther away from the Earth than the Moon. Let $d$ be the diameter of the Earth, $R_S$ be the distance to the Sun, and $R_M$ be the distance to the Moon.

Even though the Sun's mass ($M_S$) is much larger than the Moon's mass ($M_M$), the ratio of distances $(R_S/R_M)$ is very large (about 400). When cubed, this factor dominates. The Moon's much closer proximity means the difference in its gravitational pull on the near and far sides of the Earth is greater than the corresponding difference for the Sun. Therefore, the Moon's tidal effect is stronger.

(a) decreases. The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g(h) = \frac{GM_E}{(R_E+h)^2}$. As the altitude $h$ increases, the denominator increases, and thus $g(h)$ decreases.

(b) decreases. The acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g(d) = g(0)(1 - \frac{d}{R_E})$. As the depth $d$ increases, the value of $g(d)$ decreases.

(c) mass of the body. The acceleration due to gravity is given by the formula $g = \frac{GM_E}{R_E^2}$, which depends on the mass of the Earth ($M_E$) but is independent of the mass of the body ($m$) experiencing the acceleration.

(d) more. The formula $-G M m(1 / r_{2}-1 / r_{1})$ is derived directly from the universal law of gravitation, which accounts for the variation of gravitational force with distance. The formula $m g(r_{2}-r_{1})$ is an approximation that is valid only for small height differences near the Earth's surface where $g$ can be considered constant. Therefore, the first formula is more accurate.

Formula: According to Kepler's third law of planetary motion, the square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit. $T^2 \propto R^3$ Or, $\left(\frac{T_p}{T_e}\right)^2 = \left(\frac{R_p}{R_e}\right)^3$

Calculation: Substituting the given values into the formula: $\left(\frac{\frac{1}{2} T_e}{T_e}\right)^2 = \left(\frac{R_p}{R_e}\right)^3$ $\left(\frac{1}{2}\right)^2 = \left(\frac{R_p}{R_e}\right)^3$ $\frac{1}{4} = \left(\frac{R_p}{R_e}\right)^3$ Taking the cube root of both sides: $\frac{R_p}{R_e} = \left(\frac{1}{4}\right)^{1/3} = \frac{1}{4^{1/3}} \approx \frac{1}{1.587} \approx 0.63$ $R_p = R_e \times (0.25)^{1/3} \approx 0.63 R_e$

Final Answer: The orbital size of the planet would be approximately $0.63$ times the orbital size of the Earth.

Formula: From Kepler's third law, the time period $T$ of a satellite orbiting a central body of mass $M$ in an orbit of radius $R$ is given by: $T^2 = \frac{4\pi^2}{GM}R^3$ This can be rearranged to find the mass $M$: $M = \frac{4\pi^2 R^3}{G T^2}$

1. Mass of Jupiter ($M_J$): Using the data for Io: $M_J = \frac{4\pi^2 R_{Io}^3}{G T_{Io}^2}$ $M_J = \frac{4 \times (3.14)^2 \times (4.22 \times 10^8)^3}{(6.67 \times 10^{-11}) \times (1.528 \times 10^5)^2}$ $M_J = \frac{4 \times 9.86 \times 75.15 \times 10^{24}}{6.67 \times 10^{-11} \times 2.335 \times 10^{10}}$ $M_J = \frac{2964.1}{1.557} \times 10^{25} \approx 1.90 \times 10^{27} \text{ kg}$

2. Mass of the Sun ($M_S$): Using the data for Earth's orbit: $M_S = \frac{4\pi^2 R_E^3}{G T_E^2}$ $M_S = \frac{4 \times (3.14)^2 \times (1.496 \times 10^{11})^3}{(6.67 \times 10^{-11}) \times (3.156 \times 10^7)^2}$ $M_S = \frac{4 \times 9.86 \times 3.348 \times 10^{33}}{6.67 \times 10^{-11} \times 9.96 \times 10^{14}}$ $M_S = \frac{132.02}{6.643} \times 10^{29} \approx 1.987 \times 10^{30} \text{ kg}$

3. Ratio of Masses: $\frac{M_J}{M_S} = \frac{1.90 \times 10^{27} \text{ kg}}{1.987 \times 10^{30} \text{ kg}} \approx 0.000956$ $0.000956 \approx \frac{1}{1046}$ This is approximately $\frac{1}{1000}$.

Final Answer: The mass of Jupiter is calculated to be approximately $1.90 \times 10^{27} \text{ kg}$, and the mass of the Sun is approximately $1.99 \times 10^{30} \text{ kg}$. The ratio of Jupiter's mass to the Sun's mass is about $1/1046$, which is approximately one-thousandth. Hence, it is shown.

Formula: The star revolves around the galactic center, where the mass of the galaxy acts as the central mass. The time period of revolution is given by Kepler's third law: $T^2 = \frac{4\pi^2}{GM_{galaxy}}R^3$ Or, $T = \sqrt{\frac{4\pi^2 R^3}{GM_{galaxy}}}$

1. Total mass of the galaxy ($M_{galaxy}$): Assuming the mass is concentrated at the center for the star's orbit. $M_{galaxy} = n \times M_S = (2.5 \times 10^{11}) \times (2 \times 10^{30} \text{ kg}) = 5 \times 10^{41} \text{ kg}$

2. Orbital radius in meters: $R = 50,000 \text{ ly} = 5 \times 10^4 \times 9.46 \times 10^{15} \text{ m} = 4.73 \times 10^{20} \text{ m}$

3. Time period ($T$): $T = \sqrt{\frac{4 \times (3.14)^2 \times (4.73 \times 10^{20})^3}{(6.67 \times 10^{-11}) \times (5 \times 10^{41})}}$ $T = \sqrt{\frac{39.44 \times 105.7 \times 10^{60}}{33.35 \times 10^{30}}}$ $T = \sqrt{\frac{4168.3 \times 10^{60}}{33.35 \times 10^{30}}} = \sqrt{125 \times 10^{30}} = 11.18 \times 10^{15} \text{ s}$

4. Convert time period to years: 1 year $= 365.25 \times 24 \times 60 \times 60 \text{ s} = 3.156 \times 10^7 \text{ s}$ $T = \frac{11.18 \times 10^{15} \text{ s}}{3.156 \times 10^7 \text{ s/year}} \approx 3.54 \times 10^8 \text{ years}$

Final Answer: The star will take approximately $3.54 \times 10^8$ years to complete one revolution around the galactic center.

(a) kinetic energy. For a satellite in a circular orbit, the kinetic energy is $K.E. = \frac{GMm}{2r}$ and the potential energy is $P.E. = -\frac{GMm}{r}$. The total energy is $E = K.E. + P.E. = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$. Comparing the total energy and kinetic energy, we see that $E = -K.E.$. Thus, the total energy of an orbiting satellite is the negative of its kinetic energy.

(b) less. The energy required to move an object out of Earth's gravitational influence (to infinity) is the amount needed to make its total mechanical energy zero.

The formula for the escape speed from a planet of mass $M$ and radius $R$ is given by: $v_e = \sqrt{\frac{2GM}{R}}$ Let us analyze the dependencies:

(a) the mass of the body: No. The formula for escape speed does not contain the mass of the body, $m$. Therefore, the escape speed is independent of the mass of the body being projected.

(b) the location from where it is projected: Yes. The escape speed depends on the distance from the center of the Earth. The value $R$ in the formula represents this distance. Different locations on Earth's surface (e.g., poles vs. equator) have slightly different distances from the center, leading to a very small variation in escape speed.

(c) the direction of projection: No. Escape speed is a scalar quantity related to the kinetic energy required to overcome the gravitational potential energy. As long as the body is projected away from the Earth (i.e., not towards it), the direction of projection does not affect the minimum speed required to escape.

(d) the height of the location from where the body is launched: Yes. This is related to part (b). If a body is launched from a height $h$ above the Earth's surface, its distance from the center is $R_E + h$. The escape speed from that height would be $v_e = \sqrt{\frac{2GM_E}{R_E + h}}$.