JEE Physics Gravitation MCQs Set 12

Answer: (a) 11 km/s

Explanation:
1. \(v_e = \sqrt{2GM/R}\) does not contain the mass of the body.
2. So it remains 11 km/s.

Teacher's Note:
1. Escape speed is independent of the mass of the body.
2. Bigger body needs more energy, but the same speed.

Answer: (b) 44.8 km/s

Explanation:
1. \(v_e\propto\sqrt{\frac MR}\).
2. The ratio is \(\sqrt{\frac{144}{9}} = 4\), so \(v = 4\times11.2 = 44.8\) km/s.

Teacher's Note:
1. Escape speed scales with the square root of M over R.
2. Substitute the ratios directly.

Answer: (c) \(\frac{4R}{3}\)

Explanation:
1. \(\frac12v^2 = \frac18v_e^2 = \frac{GM}{4R}\).
2. Energy conservation: \(\frac{GM}{4R} = GM\left(\frac1R - \frac1r\right)\), so \(\frac1r = \frac{3}{4R}\).
3. So \(r = \frac{4R}{3}\) from the centre.

Teacher's Note:
1. Check whether the question asks for height from the centre or the surface.
2. Use energy conservation with the exact potential.

Answer: (c) \(\frac R3\)

Explanation:
1. With \(v = \frac{v_e}{2}\), energy conservation gives the highest point at distance \(\frac{4R}{3}\) from the centre.
2. The height above the surface is \(\frac{4R}{3} - R = \frac R3\).

Teacher's Note:
1. Convert distance from the centre to height above the surface.
2. Use \(\frac12v^2 = GM(\frac1R - \frac1r)\).

Answer: (c) \(\frac{R}{15}\)

Explanation:
1. \(\frac12v^2 = \frac12\cdot\frac{v_e^2}{16} = \frac{GM}{16R}\).
2. Energy conservation: \(\frac{GM}{16R} = GM\left(\frac1R - \frac1r\right)\), so \(\frac1r = \frac{15}{16R}\) and \(r = \frac{16R}{15}\).
3. The height is \(\frac{16R}{15} - R = \frac{R}{15}\).

Teacher's Note:
1. Square the 25 percent speed ratio before using energy.
2. Subtract R at the end.

Answer: (b) \(\sqrt{24}\)

Explanation:
1. \(v_e = \sqrt{2gR}\), so \(\frac{v_{earth}}{v_{moon}} = \sqrt{\frac{g_eR_e}{g_mR_m}}\).
2. With \(\frac{g_e}{g_m} = 6\) and \(\frac{R_e}{R_m} = 4\), the ratio is \(\sqrt{24}\).

Teacher's Note:
1. Combine the ratios under one square root.
2. Radius ratio equals diameter ratio.

Answer: (a) \(V_e\)

Explanation:
1. Energy conservation: \(\frac12(2V_e^2) - \frac{GM}{R} = \frac12v_\infty^2\).
2. Since \(V_e^2 = \frac{2GM}{R}\), the left side is \(\frac{2GM}{R} - \frac{GM}{R} = \frac{GM}{R}\).
3. Then \(v_\infty^2 = \frac{2GM}{R} = V_e^2\), so \(v_\infty = V_e\).

Teacher's Note:
1. Initial energy is the sum of kinetic and potential energies.
2. Leftover kinetic energy at infinity gives the speed there.

Answer: (a) \(\sqrt6 : \sqrt5\)

Explanation:
1. From infinity \(v_1^2 = \frac{2GM}{R}\).
2. From a height 5R (distance 6R) \(v_2^2 = 2GM\left(\frac1R - \frac1{6R}\right) = \frac{5}{6}\cdot\frac{2GM}{R}\).
3. So \(v_1 : v_2 = \sqrt6 : \sqrt5\).

Teacher's Note:
1. Distance from the centre is 6R when the height is 5R.
2. Apply energy conservation twice.

Answer: (b) -10368 J

Explanation:
1. The kinetic energy needed to escape equals the magnitude of the potential energy: \(\frac12mv_e^2\).
2. \(= \frac12(1)(144)^2 = 10368\) J, so \(U = -10368\) J.

Teacher's Note:
1. Total energy at escape is zero, so PE equals minus KE.
2. Square 144 carefully.

Answer: (a) 8 m/s

Explanation:
1. The escape kinetic energy is \(\frac12mv_e^2 = 64\) J.
2. So \(v_e^2 = \frac{2\times64}{2} = 64\) and \(v_e = 8\) m/s.

Teacher's Note:
1. Escape energy equals the magnitude of the surface potential energy.
2. Solve for the speed.

Answer: (c) angular speed is equal to the angular speed of rotation of the earth about its axis

Explanation:
1. A relay (geostationary) satellite appears fixed in the sky from the ground.
2. This needs the same angular speed as the earth, the same direction of rotation and an equatorial orbit.

Teacher's Note:
1. Geostationary means the same angular velocity as the earth.
2. Period 24 hours.

Answer: (d) has a time period of 24 hours and moves in the equatorial plane

Explanation:
1. To stay above one point of the earth the satellite must have the same time period as the earth, 24 hours.
2. It must also move in the equatorial plane, because only there can its orbit centre coincide with the earth's centre and match the rotation.

Teacher's Note:
1. Both conditions are needed.
2. Height is fixed by the 24 hour period.

Answer: (a) the mass of the satellite

Explanation:
1. Close to the surface, \(v = \sqrt{\frac{GM}{R}} = \sqrt{gR}\).
2. It depends on the mass and radius of the earth but not on the mass of the satellite.

Teacher's Note:
1. Orbital speed is independent of the satellite's mass.
2. The orbit radius here is the earth's radius.

Answer: (c) \(R\sqrt{\frac{g}{R+h}}\)

Explanation:
1. The orbital speed is \(v = \sqrt{\frac{GM}{R+h}}\).
2. With \(GM = gR^2\), \(v = R\sqrt{\frac{g}{R+h}}\).

Teacher's Note:
1. Replace GM by gR squared.
2. The orbital radius is R plus h.

Answer: (b) \(8\times10^3\) m/s

Explanation:
1. The least speed is the orbital velocity close to the surface: \(v = \sqrt{gR} = \sqrt{9.8\times6.4\times10^6}\).
2. This is \(7.9\times10^3\) m/s, about \(8\times10^3\) m/s.

Teacher's Note:
1. Orbital velocity near the surface is about 8 km/s.
2. Escape velocity is about 11.2 km/s.

Answer: (c) \(\sqrt{\frac23}v\)

Explanation:
1. \(v\propto\frac1{\sqrt r}\).
2. Just above the surface \(r = R\), and at altitude \(\frac R2\) we have \(r = \frac{3R}{2}\).
3. So \(v' = v\sqrt{\frac{R}{1.5R}} = \sqrt{\frac23}v\).

Teacher's Note:
1. Use the radius from the centre.
2. Speed falls as the square root of the radius increases.

Answer: (c) 1 : 2.5

Explanation:
1. \(v\propto\frac1{\sqrt r}\), so \(\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{\frac{10\times10^6}{62.5\times10^6}} = \sqrt{0.16} = 0.4\).
2. So \(v_A : v_B = 0.4 : 1 = 1 : 2.5\).

Teacher's Note:
1. Convert both radii to the same power of ten first.
2. The farther planet moves slower.

Answer: (a) \(\frac{4\pi^2}{GM}\)

Explanation:
1. Kepler's third law for a circular orbit gives \(T^2 = \frac{4\pi^2}{GM}r^3\).
2. The slope of the graph, i.e. the coefficient of \(r^3\), is \(\frac{4\pi^2}{GM}\).

Teacher's Note:
1. Derive from gravitation equals centripetal force.
2. The graph is a straight line.

Answer: (c) \(8\sqrt2\) hours

Explanation:
1. \(T\propto r^{3/2}\).
2. Near the surface \(r = R\), and at a distance 3R from the surface \(r = 4R\), so \(T' = T\times4^{3/2} = 8T\).
3. With \(T = \sqrt2\) hours, \(T' = 8\sqrt2\) hours (about 11.3 hours).

Teacher's Note:
1. Add R to 3R to get the radius of 4R.
2. 4 to the power 3/2 is 8.

Answer: (a) 1 : 1

Explanation:
1. \(T^2 = \frac{4\pi^2r^3}{GM}\) does not contain the mass of the satellite.
2. The same period therefore means the same radius, ratio 1 : 1.

Teacher's Note:
1. Orbit radius depends on the period, not the satellite mass.
2. Geostationary orbits all have the same radius.

Answer: (c) 40 hours

Explanation:
1. \(T\propto r^{3/2}\), so \(T' = 5\times4^{3/2} = 5\times8\).
2. The new period is 40 hours.

Teacher's Note:
1. Period varies as the 3/2 power of the radius.
2. 4 to the power 3/2 is 8.

Answer: (c) 27 days

Explanation:
1. A communication satellite has a period of 1 day.
2. \(T\propto r^{3/2}\), so \(T_S = 1\times9^{3/2} = 27\) days.

Teacher's Note:
1. A communication satellite is geostationary with T equal to 1 day.
2. 9 to the power 3/2 is 27.

Answer: (c) \(\left[\frac{GMT^2}{4\pi^2}\right]^{1/3}\)

Explanation:
1. The satellite must have period T, and gravity supplies the centripetal force: \(\frac{GMm}{r^2} = m\omega^2r\) with \(\omega = \frac{2\pi}{T}\).
2. So \(r^3 = \frac{GMT^2}{4\pi^2}\).
3. The question asks for the distance from the earth's centre, so \(r = \left[\frac{GMT^2}{4\pi^2}\right]^{1/3}\).

Teacher's Note:
1. Distance from the centre does not need the extra R.
2. Use the 24 hour period.

Answer: (d) \(8\times2\pi\sqrt{\frac Rg}\)

Explanation:
1. \(T = 2\pi\sqrt{\frac{r^3}{GM}}\) with \(GM = gR^2\) and \(r = 4R\).
2. \(T = 2\pi\sqrt{\frac{64R^3}{gR^2}} = 8\times2\pi\sqrt{\frac Rg}\).

Teacher's Note:
1. Replace GM by gR squared.
2. 64 under the root gives 8.

Answer: (c) \(-\frac{GMm}{R}\)

Explanation:
1. The gravitational potential energy of the pair is \(U = -\frac{GMm}{R}\), taking zero at infinity.
2. This is true for any orbit radius R.

Teacher's Note:
1. Potential energy is negative.
2. Do not confuse U with the total energy.

Answer: (c) 3.28 km/s

Explanation:
1. The orbital speed near the surface is \(v_0 = 7.9\) km/s and the escape speed is \(\sqrt2v_0 = 11.2\) km/s.
2. The additional velocity is \(11.2 - 7.9 = 3.3\) km/s, i.e. \((\sqrt2 - 1)\times7.9\approx3.28\) km/s.

Teacher's Note:
1. Escape speed is root two times the orbital speed.
2. Subtract the speed already present.