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}

Gravitational Field Intensity — JEE Main Physics MCQs with Solutions

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