Question 1

If the gravitational field intensity on the surface of the Earth is I_0, what is the magnitude of the gravitational field intensity at a distance of 2R from the surface of the Earth? (R = radius of the Earth)

Accepted Answer

\dfrac{I_0}{9}. Distance from the centre of the Earth: r = R + 2R = 3R.
Outside the Earth (r \ge R): I(r) = \frac{GM}{r^2} = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} = \frac{I_0}{9}

Question 2

At what height h above the surface of the Earth will the gravitational field intensity reduce to 16\% of its value on the surface? (R = radius of the Earth)

Accepted Answer

1.5\,R. Given I(h) = 0.16\,I_0 = \dfrac{4}{25}\,I_0.
Using I(h) = I_0\left(\dfrac{R}{R+h}\right)^2: \left(\frac{R}{R+h}\right)^2 = \frac{4}{25} \implies \frac{R}{R+h} = \frac{2}{5} \implies 5R = 2R + 2h \implies h = 1.5\,R

Question 3

Two satellites A and B are orbiting the Earth (radius R) in circular orbits at heights h_A = R and h_B = 3R above the surface, respectively. What is the ratio of the gravitational field intensity experienced by satellite A to that by satellite B, \dfrac{I_A}{I_B}?

Accepted Answer

4 : 1. Distance of A from the centre: r_A = R + R = 2R.
Distance of B from the centre: r_B = R + 3R = 4R.
Since I \propto \dfrac{1}{r^2}: \frac{I_A}{I_B} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{4R}{2R}\right)^2 = 4 : 1

Question 4

If a body is moved radially outward from a distance r_1 = 2R to r_2 = 4R measured from the centre of the Earth (radius R), the percentage decrease in the gravitational field intensity is:

Accepted Answer

75\%. Field at r_1 = 2R: I_1 = \dfrac{GM}{(2R)^2} = \dfrac{I_0}{4}.
Field at r_2 = 4R: I_2 = \dfrac{GM}{(4R)^2} = \dfrac{I_0}{16}.
Percentage decrease: \frac{I_1 - I_2}{I_1} 	imes 100\% = \frac{\frac{1}{4} - \frac{1}{16}}{\frac{1}{4}} 	imes 100\% = 75\%

Question 5

The gravitational field intensity on the surface of the Earth (radius R, assumed a uniform solid sphere) is I_0. What is the magnitude of the gravitational field intensity inside the Earth at a depth of d = \dfrac{R}{2} from the surface?

Accepted Answer

\dfrac{I_0}{2}. Distance from the centre: r = R - \dfrac{R}{2} = \dfrac{R}{2}.
Inside a uniform solid sphere (r \le R): I(r) = \dfrac{GM}{R^3}\,r = I_0\left(\dfrac{r}{R}\right) I = I_0\left(\frac{R/2}{R}\right) = \frac{I_0}{2}

Question 6

A straight mine shaft is drilled deep into the Earth (radius R, assumed a uniform solid sphere, surface field intensity I_0). At what depth d below the surface does the gravitational field intensity become \dfrac{1}{3}I_0?

Accepted Answer

\dfrac{2R}{3}. Inside the Earth: I(d) = I_0\left(1 - \dfrac{d}{R}\right).
Setting I(d) = \dfrac{1}{3}I_0: 1 - \frac{d}{R} = \frac{1}{3} \implies \frac{d}{R} = \frac{2}{3} \implies d = \frac{2R}{3}

Question 7

Consider two points P_1 and P_2: P_1 is located at a depth of \dfrac{R}{4} below the surface of the Earth, and P_2 is at an altitude of \dfrac{R}{4} above the surface (R = radius of the Earth, assumed a uniform solid sphere). The ratio of the magnitude of gravitational field intensity at P_1 to that at P_2 is:

Accepted Answer

75 : 64. At P_1 (inside, depth R/4): I_1 = I_0\left(1 - \dfrac{1}{4}ight) = \dfrac{3}{4}I_0.
At P_2 (outside, height R/4): I_2 = I_0\left(\dfrac{R}{R + R/4}ight)^2 = I_0\left(\dfrac{4}{5}ight)^2 = \dfrac{16}{25}I_0.
Ratio: \frac{I_1}{I_2} = \frac{3/4}{16/25} = \frac{3 	imes 25}{4 	imes 16} = \frac{75}{64}

Question 8

A thin, uniform spherical shell of mass M and radius R is centred at the origin. Let r be the radial distance from the centre. Which of the following statements is CORRECT regarding the gravitational field intensity \vec{I} and its magnitude I(r)?

Accepted Answer

Inside the shell (r < R), \vec{I} = \vec{0}; outside the shell (r > R), the magnitude I(r) = \dfrac{GM}{r^2}. By Newton's shell theorem: • Inside (r < R): the pulls of symmetric mass elements cancel in pairs, so \vec{I}_{ ext{in}} = \vec{0}. • Outside (r > R): the shell acts as if all its mass were at the centre, so \vec{I}_{ ext{out}} = -\dfrac{GM}{r^2}\hat{r}, with magnitude I = \dfrac{GM}{r^2}.

Question 9

A thin, uniform circular ring of mass M and radius R lies in the y-z plane centred at the origin. What is the magnitude of the gravitational field intensity at an axial distance x = \sqrt{2}R from the centre of the ring?

Accepted Answer

\dfrac{\sqrt{2}}{3\sqrt{3}}\,\dfrac{GM}{R^2}. On the axis of a ring: I(x) = \dfrac{GMx}{(R^2 + x^2)^{3/2}}.
With x = \sqrt{2}R: R^2 + x^2 = 3R^2, so (3R^2)^{3/2} = 3\sqrt{3}\,R^3. I = \frac{GM(\sqrt{2}R)}{3\sqrt{3}\,R^3} = \frac{\sqrt{2}}{3\sqrt{3}}\,\frac{GM}{R^2}

Question 10

The gravitational field intensity at a distance \dfrac{R}{2} from the centre of the Earth is I_0. What will be the magnitude of the gravitational field intensity at a distance 2R from the centre? (Assume the Earth is a uniform solid sphere of radius R)

Accepted Answer

\dfrac{I_0}{2}. Inside a uniform solid sphere (r \le R): I(r) = \dfrac{GM}{R^3}\,r.
At r = \dfrac{R}{2}: I_0 = \dfrac{GM}{2R^2} \implies \dfrac{GM}{R^2} = 2I_0.
Outside (r \ge R): I(r) = \dfrac{GM}{r^2}, so at r = 2R: I(2R) = \frac{GM}{4R^2} = \frac{1}{4}(2I_0) = \frac{I_0}{2}

Question 11

What will be the gravitational field intensity on the surface of a planet whose mass is 4 times and radius is half that of the Earth? (I_e = gravitational field intensity on the surface of the Earth)

Accepted Answer

16\,I_e. On the Earth: I_e = \dfrac{GM_e}{R_e^2}.
For the planet, M_p = 4M_e and R_p = \dfrac{R_e}{2}: I_p = \frac{G(4M_e)}{(R_e/2)^2} = 16\,\frac{GM_e}{R_e^2} = 16\,I_e

Question 12

A hypothetical celestial body has a mass 81 times that of the Earth, and the distance between their centres is 4	imes 10^5	ext{ km}. Find the distance from the Earth where the net gravitational field intensity due to the two is zero.

Accepted Answer

40{,}000	ext{ km}. Let the neutral point be at distance x from the Earth; its distance from the other body is d - x, with d = 4	imes 10^5 km.
\frac{GM_E}{x^2} = \frac{G(81M_E)}{(d - x)^2} \implies \frac{1}{x} = \frac{9}{d - x} \implies d = 10x x = \frac{d}{10} = 40{,}000	ext{ km}

Question 13

The gravitational field intensity on the surface of the Earth (radius R, assumed a uniform solid sphere) is I_s. At what distances r from the centre of the Earth (one inside and one outside) is the magnitude of the field intensity equal to \dfrac{I_s}{4}?

Accepted Answer

Inside: \dfrac{R}{4}, Outside: 2R. Inside (r < R): I(r) = I_s\left(\dfrac{r}{R}\right) = \dfrac{I_s}{4} \implies r = \dfrac{R}{4}.
Outside (r > R): I(r) = I_s\left(\dfrac{R}{r}\right)^2 = \dfrac{I_s}{4} \implies \dfrac{R}{r} = \dfrac{1}{2} \implies r = 2R.

Question 14

A spherical planet has the same average mass density as the Earth, but its radius is 3 times that of the Earth. If the gravitational field intensity on the Earth's surface is I_e, what is the field intensity on the surface of this planet?

Accepted Answer

3\,I_e. In terms of density \rho: I = \frac{GM}{R^2} = \frac{G\left(\frac{4}{3}\pi R^3\rho\right)}{R^2} = \frac{4}{3}\pi G\rho R With \rho the same, I \propto R, so I_p = 3\,I_e.

Question 15

Two point masses M_1 = 4M and M_2 = 16M are held at a fixed separation L. At what distance from M_1 along the line joining the two masses is the net gravitational field intensity equal to zero?

Accepted Answer

\dfrac{L}{3}. Let the point be at distance x from M_1; its distance from M_2 is L - x.
\frac{G(4M)}{x^2} = \frac{G(16M)}{(L - x)^2} \implies \frac{2}{x} = \frac{4}{L - x} \implies L - x = 2x \implies x = \frac{L}{3}

Question 16

The time period of a geo-stationary satellite is 24 h, at a height 6R_E (R_E is the radius of earth) from surface of earth. The time period of another satellite whose height is 2.5R_E from surface will be

Accepted Answer

6\sqrt{2} h. From Kepler's third law, the time period of revolution of satellite around earth is T^2 \propto r^3 or T \propto r^{3/2}, where r is the radius of the satellite's orbit. Here, r_1 = 6R_E + R_E, T_1 = 24 h and r_2 = 2.5R_E + R_E. So, \dfrac{T_1}{T_2} = \left(\dfrac{r_1}{r_2}\right)^{3/2}, \dfrac{24}{T_2} = \left(\dfrac{6R_E + R_E}{2.5R_E + R_E}\right)^{3/2} = \left(\dfrac{7}{3.5}\right)^{3/2}. \Rightarrow T_2 = \dfrac{24}{(2)^{3/2}} = \dfrac{24}{2\sqrt{2}} = \dfrac{12}{\sqrt{2}} = 6\sqrt{2} h.

Question 17

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A, K_B and K_C, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

Accepted Answer

K_A > K_B > K_C. The closed point A is the perihelion and the farthest point C is the aphelion. As per Kepler's second law of areas, the planet moves slowly (v_{min}) when farthest from the Sun and more rapidly (v_{max}) when nearest to the Sun. Thus v_A = v_{max}, v_C = v_{min}, so v_A > v_B > v_C. Kinetic energy K = \dfrac{1}{2}mv^2, hence K_A > K_B > K_C.

Question 18

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between the sun and planet i.e. T^2 = Kr^3, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitation force of attraction between them is F = \dfrac{GMm}{r^2}, here G is gravitational constant. The relation between G and K is described as

Accepted Answer

GMK = 4\pi^2. The gravitational force of attraction between the planet and sun provides the centripetal force, i.e. \dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{\dfrac{GM}{r}}. The time period of the planet is T = \dfrac{2\pi r}{v} \Rightarrow T^2 = \dfrac{4\pi^2 r^2}{GM/r} = \dfrac{4\pi^2 r^3}{GM} ...(i). Also from Kepler's third law T^2 = Kr^3 ...(ii). From Eqs. (i) and (ii), \dfrac{4\pi^2 r^3}{GM} = Kr^3 \Rightarrow GMK = 4\pi^2.

Question 19

Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

Accepted Answer

7.5R. Suppose the smaller body (mass M) covers a distance x before collision. The gap between the surfaces is 12R - R - 2R = 9R. Since the centre of mass stays fixed, Mx = 5M(9R - x), or x = 45R - 5x, or x = \dfrac{45R}{6} = 7.5R.

Question 20

A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hour at a height of 2R from the surface of the earth is

Accepted Answer

6\sqrt{2}. From Kepler's third law T^2 \propto r^3, where r is the radius of the orbit (semi-major axis). Hence T_1^2 \propto r_1^3 and T_2^2 \propto r_2^3. So \dfrac{T_2^2}{T_1^2} = \dfrac{r_2^3}{r_1^3} = \dfrac{(3R)^3}{(6R)^3} = \dfrac{1}{8}. 	herefore T_2^2 = \dfrac{1}{8}T_1^2 \Rightarrow T_2 = \dfrac{24}{2\sqrt{2}} = 6\sqrt{2} h (taking T_1 = 24 h).

Gravitation — JEE Main Physics MCQs with Solutions

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