JEE Physics Gravitation MCQs Set 20

Answer: (b) \( 4.8\times10^{-5}\text{ N} \)

Explanation:
1. Use \( F=\dfrac{GMm}{r^2} \) with \( r=3.78\times10^{8}\text{ m} \).
2. \( F=\dfrac{6.67\times10^{-11}\times7.34\times10^{22}\times1.4}{(3.78\times10^{8})^2}\approx4.8\times10^{-5}\text{ N} \).

Teacher's Note:
1. The pull of the moon on a body of ordinary size is tiny compared with its weight of about 14 N.
2. Always convert the distance to metres before using the formula.

Answer: (d) \( -20\hat i-40\hat j\text{ N/kg};\ 5\sqrt5\text{ N} \)

Explanation:
1. The field is \( \vec E=-\nabla V=-\dfrac{\partial V}{\partial x}\hat i-\dfrac{\partial V}{\partial y}\hat j=-20\hat i-40\hat j \), the same at every point.
2. Magnitude \( =\sqrt{20^2+40^2}=20\sqrt5\text{ N/kg} \).
3. Force \( =0.25\times20\sqrt5=5\sqrt5\approx11.2\text{ N} \).

Teacher's Note:
1. The field points towards decreasing potential, which is why the negative sign appears.
2. A potential that is linear in x and y gives a uniform field.

Answer: (c) \( -6.3\times10^{7}\text{ J/kg} \)

Explanation:
1. \( V=-\dfrac{GM}{R}=-\dfrac43\pi G\rho R^2 \).
2. \( V=-\dfrac43\pi(6.67\times10^{-11})(5.5\times10^3)(6.4\times10^6)^2\approx-6.3\times10^7\text{ J/kg} \).

Teacher's Note:
1. Writing the mass as density times volume lets you use the density given in the question.
2. The potential at the surface is negative because it is measured from zero at infinity.

Answer: (a) \( 2\sqrt{\dfrac{G(M_A+M_B)}{d}} \)

Explanation:
1. Each planet is at distance \( d/2 \) from the midpoint, so the potential per unit mass there is \( -\dfrac{2G(M_A+M_B)}{d} \).
2. For escape, \( \tfrac12v^2=\dfrac{2G(M_A+M_B)}{d} \), giving \( v=2\sqrt{\dfrac{G(M_A+M_B)}{d}} \).

Teacher's Note:
1. Potentials add as plain numbers, so there is no cancellation like that of forces.
2. The satellite just reaches infinity with zero speed when its total energy is zero.

Answer: (d) 1.5%

Explanation:
1. \( T\propto r^{3/2} \), so \( \dfrac{T_2}{T_1}=(1.01)^{3/2}\approx1.015 \).
2. The period is about 1.5% larger.

Teacher's Note:
1. For small changes, \( \dfrac{\Delta T}{T}=\dfrac32\dfrac{\Delta r}{r} \).
2. This follows directly from Kepler's third law.

Answer: (b) \( \dfrac{GM^2}{4R} \)

Explanation:
1. Potential energy of the pair \( =-\dfrac{GM^2}{2R} \).
2. Kinetic energy of the two stars \( =2\times\tfrac12M\cdot\dfrac{GM}{4R}=\dfrac{GM^2}{4R} \).
3. Total energy \( =-\dfrac{GM^2}{4R} \), so the energy needed to make it zero is \( \dfrac{GM^2}{4R} \).

Teacher's Note:
1. The system is bound, so its total energy is negative.
2. The energy to unbind a system equals the magnitude of its total energy.

Answer: (a) \( 2\sqrt{\dfrac{GM}{R}} \)

Explanation:
1. The centre of mass is at distance \( R \) from each star, so the potential per unit mass there is \( -\dfrac{2GM}{R} \).
2. Escape needs \( \tfrac12v^2=\dfrac{2GM}{R}\Rightarrow v=2\sqrt{\dfrac{GM}{R}} \).

Teacher's Note:
1. The meteorite feels no net force at the centre of mass, but it still has to climb out of the combined potential well.
2. Add the two potentials, then set the total energy to zero.

Answer: (c) \( 9.4\text{ m/s}^2 \)

Explanation:
1. At height \( R/20 \): \( g'=g\left(\dfrac{20}{21}\right)^2=9\Rightarrow g=9\times\left(\dfrac{21}{20}\right)^2=9.92\text{ m/s}^2 \).
2. At depth \( R/20 \): \( g''=g\left(1-\dfrac1{20}\right)=9.92\times0.95\approx9.4\text{ m/s}^2 \).

Teacher's Note:
1. Above the surface gravity falls as the inverse square of the distance from the centre; below the surface it falls linearly with depth.
2. First find the surface value of g, then apply the depth formula.

Answer: (b) 0.998 s

Explanation:
1. At the equator the effective gravity is \( g_e=g-\omega^2R \); at the pole it is \( g \).
2. With \( \omega=\dfrac{2\pi}{86400}\text{ rad/s} \), \( \dfrac{\omega^2R}{g}\approx3.4\times10^{-3} \).
3. Since \( T\propto\dfrac1{\sqrt{g}} \), \( T_{pole}=1.00\times\sqrt{\dfrac{g-\omega^2R}{g}}\approx\sqrt{0.9966}\approx0.998\text{ s} \).

Teacher's Note:
1. The pendulum is faster at the pole because gravity is stronger there.
2. The effect of rotation is small, only a few parts in a thousand.

Answer: (d) \( \sqrt{\dfrac{GM}{L}} \)

Explanation:
1. The orbit radius is the circumradius \( r=\dfrac{L}{\sqrt3} \).
2. Net force on a star \( =2\cdot\dfrac{GM^2}{L^2}\cos30^\circ=\dfrac{\sqrt3\,GM^2}{L^2} \).
3. Setting \( \dfrac{Mv^2}{r}=\dfrac{\sqrt3GM^2}{L^2} \) gives \( v^2=\dfrac{\sqrt3GM}{L^2}\cdot\dfrac{L}{\sqrt3}=\dfrac{GM}{L} \).

Teacher's Note:
1. Do not use the side of the triangle as the radius of the circle.
2. The two pulls make an angle of 60 degrees, so their resultant points to the centre.

Answer: (c) \( -2Gm \)

Explanation:
1. \( V=-Gm\left(1+\dfrac12+\dfrac14+\dfrac18+\cdots\right) \).
2. The geometric series with first term 1 and ratio \( \tfrac12 \) sums to 2, so \( V=-2Gm \).

Teacher's Note:
1. Potential is a scalar, so the contributions are simply added.
2. The infinite geometric series sums to first term over (1 - ratio).

Answer: (a) \( \dfrac{3Gm^2}{2a} \)

Explanation:
1. Potential energy of the system \( U=-\dfrac{3Gm^2}{a} \) initially and \( -\dfrac{3Gm^2}{2a} \) finally.
2. Work done by the external agent \( =U_f-U_i=\dfrac{3Gm^2}{a}-\dfrac{3Gm^2}{2a}=\dfrac{3Gm^2}{2a} \).

Teacher's Note:
1. Pulling attractive masses apart needs positive work from outside.
2. There are three pairs, each contributing \( -\dfrac{Gm^2}{\text{side}} \).

Answer: (d) \( 2\sqrt{\dfrac{Gm}{a}} \)

Explanation:
1. Initial potential energy of the moving mass \( =-\dfrac{2Gm^2}{a} \); at the midpoint (distance \( a/2 \) from each) \( =-\dfrac{4Gm^2}{a} \).
2. Gain in kinetic energy \( \tfrac12mv^2=\dfrac{2Gm^2}{a} \), so \( v=2\sqrt{\dfrac{Gm}{a}} \).

Teacher's Note:
1. The moving mass slides along the perpendicular bisector, so it reaches the midpoint.
2. The mutual energy of the two fixed masses does not change.

Answer: (b) 11 km/s

Explanation:
1. Escape velocity follows from energy conservation, \( \tfrac12v^2=\dfrac{GM}{R} \), which involves only the speed.
2. It does not depend on the direction of projection, so it stays 11 km/s.

Teacher's Note:
1. Gravity is conservative, so only the initial speed and position matter for escape.
2. Air resistance and the earth's rotation are neglected.

Answer: (a) \( \dfrac{7GMm}{36R^2} \)

Explanation:
1. Full sphere: \( F_1=\dfrac{GMm}{(2R)^2}=\dfrac{GMm}{4R^2} \).
2. The removed sphere has mass \( M/8 \) and its centre is at distance \( 2R-\tfrac R2=\tfrac{3R}{2} \) from \( m \): \( F_2=\dfrac{G(M/8)m}{(3R/2)^2}=\dfrac{GMm}{18R^2} \).
3. Net force \( =F_1-F_2=\dfrac{GMm}{R^2}\left(\dfrac14-\dfrac1{18}\right)=\dfrac{7GMm}{36R^2} \).

Teacher's Note:
1. Use superposition: the hollowed sphere equals the full sphere plus a sphere of negative mass in the cavity.
2. Mass scales with the cube of the radius, so the cavity has one eighth of the mass.

Answer: (c) is the same everywhere

Explanation:
1. Let \( \rho=\dfrac kr \). Mass within \( r \): \( \int_0^r4\pi x^2\dfrac kx\,dx=2\pi kr^2 \).
2. Field \( g=\dfrac{G\cdot2\pi kr^2}{r^2}=2\pi Gk \), independent of \( r \).

Teacher's Note:
1. Check the limiting case: uniform density gives a field proportional to r.
2. Here the density falls just fast enough to make the field constant.

Answer: (b) \( \tfrac34mgR \)

Explanation:
1. \( \Delta U=GMm\left(\dfrac1R-\dfrac1{4R}\right)=\dfrac{3GMm}{4R} \).
2. Since \( GM=gR^2 \), \( \Delta U=\tfrac34mgR \).

Teacher's Note:
1. The formula mgh would give 3mgR, which is wrong for heights comparable with the radius.
2. The distance from the centre after lifting is R + 3R = 4R.

Answer: (d) \( \dfrac{R}{1-k^2} \)

Explanation:
1. Energy: \( \tfrac12k^2v_e^2-\dfrac{GM}{R}=-\dfrac{GM}{r} \) and \( \tfrac12v_e^2=\dfrac{GM}{R} \).
2. So \( \dfrac{GM}{R}(k^2-1)=-\dfrac{GM}{r}\Rightarrow r=\dfrac{R}{1-k^2} \).

Teacher's Note:
1. As k approaches 1 the distance becomes infinite, as expected for escape.
2. The answer is a distance from the centre, not a height above the surface.

Answer: (c) \( \dfrac{r}{4^{1/3}} \)

Explanation:
1. A geostationary satellite has the planet's rotational period \( T \), and \( T^2\propto r^3 \).
2. The period halves, so \( r'^3=\dfrac{r^3}{4}\Rightarrow r'=\dfrac{r}{4^{1/3}} \).

Teacher's Note:
1. If the planet spins faster, the stationary orbit moves closer.
2. Use the cube root, not the square root.

Answer: (a) \( \sqrt{\dfrac{gR^2}{R+x}} \)

Explanation:
1. \( \dfrac{mv^2}{R+x}=\dfrac{GMm}{(R+x)^2}\Rightarrow v=\sqrt{\dfrac{GM}{R+x}} \).
2. With \( GM=gR^2 \): \( v=\sqrt{\dfrac{gR^2}{R+x}} \).

Teacher's Note:
1. Replace GM by gR^2 to express the answer in terms of g.
2. The orbit radius is R + x.

Answer: (d) The second law (constant areal velocity)

Explanation:
1. Any force directed along the line joining planet and sun is central and gives zero torque about the sun.
2. So angular momentum, and hence areal velocity, stays constant.
3. The elliptical orbit and the exponent in the period law both depend on the inverse square form.

Teacher's Note:
1. Areal velocity equals half the angular momentum per unit mass.
2. The second law is true for every central force.