JEE Physics Gravitation MCQs Set 32

Answer: (d) 1

Explanation:
1. \( KE=\dfrac{GMm}{2r} \). \( A \): \( \dfrac{GMm}{2R} \). \( B \): \( \dfrac{GM(2m)}{2(2R)}=\dfrac{GMm}{2R} \).
2. The ratio is 1.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( mv^2 \)

Explanation:
1. Escape speed at the orbit \( =\sqrt2\,v \).
2. \( KE=\tfrac12m(\sqrt2v)^2=mv^2 \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (b) \( 3.2\times10^3 \) km

Explanation:
1. \( E_1=GMm\left(\dfrac1R-\dfrac1{R+h}\right)=\dfrac{GMmh}{R(R+h)} \). \( E_2=\dfrac{GMm}{2(R+h)} \).
2. Equal when \( \dfrac hR=\dfrac12\Rightarrow h=\dfrac R2=3.2\times10^3 \) km.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (c) \( \dfrac{5GmM}{6R} \)

Explanation:
1. Orbit radius \( 3R \): total energy \( =-\dfrac{GMm}{6R} \). Initial energy at the surface \( =-\dfrac{GMm}R \).
2. Required \( =-\dfrac{GMm}{6R}+\dfrac{GMm}R=\dfrac{5GMm}{6R} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (d) 2 h

Explanation:
1. \( T^2\propto r^3 \): \( T=24\left(\dfrac{6400+\text{few hundred}}{36000}\right)^{3/2} \).
2. With \( r\approx6700 \) km: \( T\approx24\times0.08\approx1.9 h \), about \( 2 h \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) The acceleration of S is always directed towards the centre of the earth

Explanation:
1. Gravity is a central force, so the acceleration always points to the centre of the earth.
2. Angular momentum is conserved in both magnitude and direction; total energy is constant; the speed (hence linear momentum) varies along the ellipse.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Statement I is true: the astronaut feels no normal reaction.
2. Statement II is true and explains it: station and astronaut both fall freely with the same acceleration, so there is no contact force between them.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (c) \( 6\sqrt2\approx8.5 \)

Explanation:
1. \( T^2\propto r^3 \): \( T=24\left(\dfrac{3.5R}{7R}\right)^{3/2}=24\times\dfrac1{2\sqrt2} \).
2. \( T=6\sqrt2\approx8.5 h \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (d) False, because only a satellite orbiting in the equatorial plane can be geostationary

Explanation:
1. The centre of the orbit must be the centre of the earth, so the orbital plane passes through it.
2. To stay over one point the satellite must have a \( 24 h \) period in the equatorial plane; New Delhi is not on the equator, so this is impossible.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) 6400 km

Explanation:
1. \( \sqrt{\dfrac{GM}r}=\dfrac12\sqrt{\dfrac{2GM}R}\Rightarrow\dfrac1r=\dfrac1{2R}\Rightarrow r=2R \).
2. Height \( =r-R=R=6400 \) km.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (b) 7.9 km/s

Explanation:
1. \( \tfrac12v^2=GM\left(\dfrac1R-\dfrac1{2R}\right)=\dfrac{GM}{2R}\Rightarrow v=\sqrt{\dfrac{GM}R}=\sqrt{gR} \).
2. \( v=\sqrt{9.8\times6.4\times10^6}=7.9 \) km/s.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (c) \( \pi\times10^4 \)

Explanation:
1. \( T^2\propto r^3 \): \( r_2=r_1(8)^{2/3}=4\times10^4 \) km.
2. \( v_1=\dfrac{2\pi(10^4)}1=2\pi\times10^4 \) km/h and \( v_2=\dfrac{2\pi(4\times10^4)}8=\pi\times10^4 \) km/h.
3. When closest, both move in the same direction: relative speed \( =v_1-v_2=\pi\times10^4 \) km/h.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (d) \( 3\times10^{-4} \) rad/s

Explanation:
1. Relative speed \( =\pi\times10^4 \) km/h \( =8.73 \) km/s, perpendicular to the line joining them.
2. \( \omega=\dfrac{8.73}{3\times10^4}\approx2.9\times10^{-4}\approx3\times10^{-4} \) rad/s.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( \dfrac{L}{2m} \)

Explanation:
1. \( \dfrac{dA}{dt}=\dfrac12r^2\omega=\dfrac12r v=\dfrac L{2m} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (b) \( T\propto R^{\frac{n+1}2} \)

Explanation:
1. \( \dfrac{mv^2}R\propto R^{-n}\Rightarrow v^2\propto R^{1-n} \).
2. \( T=\dfrac{2\pi R}v\propto R\cdot R^{\frac{n-1}2}=R^{\frac{n+1}2} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (c) \( T_A=T_B \)

Explanation:
1. The two stars of a binary system orbit their common centre of mass and stay on opposite sides of it.
2. So they complete one revolution in the same time: \( T_A=T_B \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (d) 129

Explanation:
1. \( T^2\propto r^3\Rightarrow T=365\left(\dfrac12\right)^{3/2}=\dfrac{365}{2.83} \).
2. \( T\approx129 \) days.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( T^2\propto R^{7/2} \)

Explanation:
1. \( \dfrac{mv^2}R\propto R^{-5/2}\Rightarrow v^2\propto R^{-3/2} \).
2. \( T^2=\dfrac{4\pi^2R^2}{v^2}\propto R^{2+3/2}=R^{7/2} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (b) \( 6.94\times10^{22} \) m²

Explanation:
1. Areal velocity \( =\dfrac L{2m}=2.2\times10^{15} m^{2}/s \).
2. In one year (\( 3.156\times10^7 s \)) the radius vector sweeps the whole orbit: \( A=2.2\times10^{15}\times3.156\times10^7\approx6.94\times10^{22} m^{2} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (c) angular momentum

Explanation:
1. The gravitational force is central, so the torque about the sun is zero.
2. Hence angular momentum is conserved, and \( \dfrac{dA}{dt}=\dfrac L{2m} \) is constant.

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (d) \( 1.16\sqrt{\dfrac{GM}a} \)

Explanation:
1. Orbit radius \( r=\dfrac a{\sqrt2} \). Net inward force on one particle: \( 2\dfrac{GM^2}{a^2}\cos45^\circ+\dfrac{GM^2}{(\sqrt2a)^2}=\dfrac{GM^2}{a^2}\left(\sqrt2+\tfrac12\right) \).
2. \( \dfrac{Mv^2}r=\) force \( \Rightarrow v^2=\dfrac{GM}a\left(1+\dfrac1{2\sqrt2}\right)=1.354\dfrac{GM}a \).
3. \( v=1.16\sqrt{\dfrac{GM}a} \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( 2\sqrt3 \) s

Explanation:
1. \( g\propto\dfrac M{R^2} \): \( g_p=\dfrac3{9}g_e=\dfrac{g_e}3 \).
2. \( T=2\pi\sqrt{\dfrac lg}\propto\dfrac1{\sqrt g}\Rightarrow T_p=2\sqrt3 s \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.