NEET Gravitation MCQs & PYQs

Practice NEET Gravitation Questions

Question 1:

moderate

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 sun and planet, i.e., $T^2 = Kr^3$ here $K$ is constant. If the masses of sun and planet are $M$ and $m$ respectively then as per Newton’s law of gravitation force of attraction between them is $F = \frac{GMm}{r^2}$ here $G$ is gravitational constant. The relation between $G$ and $K$ is described as:

(2015)

We know that the time period of a planet is given by $T^2 = \frac{4\pi^2}{GM}r^3$. Comparing this with $T^2 = Kr^3$, we get $K = \frac{4\pi^2}{GM}$. Rearranging this gives $GMK = 4\pi^2$.

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Question 2:

moderate

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 hours at a height of $2R$ from the surface of the earth is:

(2012 Pre)

For the geostationary satellite, $T_1 = 24\text{ hours}$, $r_1 = R + 5R = 6R$. For the second satellite, $r_2 = R + 2R = 3R$. Using Kepler's third law $T^2 \propto r^3$, we have $T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{6R}\right)^{3/2} = 24 \left(\frac{1}{2}\right)^{3/2} = 6\sqrt{2}\text{ hours}$.

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Question 3:

moderate

A planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively, then the ratio is

(2011 Mains)

By the conservation of angular momentum at the closest and farthest points, $mv_1r_1 = mv_2r_2$. Therefore, the ratio of their linear velocities $\frac{v_1}{v_2}$ is equal to $\frac{r_2}{r_1}$.

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Question 4:

moderate

The period of revolution of planet A around the sun is $8$ times that of B. The distance of A from the sun is how many times greater than that of B from the sun?

(1997)

According to Kepler's third law, $T^2 \propto r^3$. Given $T_A = 8T_B$, so $\left(\frac{r_A}{r_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2 = (8)^2 = 64$. Taking the cube root yields $r_A = 4r_B$.

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Question 5:

moderate

The distance of two planets from the sun are $10^{13}\text{ m}$ and $10^{12}\text{ m}$ respectively. The ratio of time periods of the planets is:

(1994, 88)

Using Kepler's 3rd Law $T^2 \propto r^3$, the ratio of time periods is $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2}$. Substituting the distances, we get $\frac{T_1}{T_2} = \left(\frac{10^{13}}{10^{12}}\right)^{3/2} = (10)^{3/2} = 10\sqrt{10}$.

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Question 6:

moderate

A satellite A of mass $m$ is at a distance of $r$ from the surface of the earth. Another satellite B of mass $2m$ is at a distance of $2r$ from the earth’s centre. Their time periods are in the ratio of:

(1993)

Time period of a satellite is independent of its mass. Assuming the distance $r$ for satellite A was intended to be from the centre (a standard typo in this question format), $r_A = r$ and $r_B = 2r$. Their time period ratio $\frac{T_A}{T_B} = \left(\frac{r_A}{r_B}\right)^{3/2} = \left(\frac{1}{2}\right)^{3/2} = 1:2\sqrt{2}$.

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Question 7:

moderate

The largest and the shortest distance of the earth from the sun are $r_1$ and $r_2$. Its distance from the sun when it is at perpendicular to the major axis of the orbit drawn from the sun is:

(1988)

The distance from the sun when the planet is perpendicular to the major axis drawn from the sun is the semi-latus rectum of the elliptical orbit. It is calculated as the harmonic mean of the apoapsis and periapsis distances, giving $\frac{2r_1r_2}{r_1+r_2}$.

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Question 8:

moderate

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

(2018)

The value of acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$. If $G$ increases by $10$ times, $g$ also increases by $10$ times since it depends on the mass of the Earth, not the Sun. Hence, the statement that '$g$' will not change is incorrect.

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Question 9:

difficult

Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is:

(1995)

The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.

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Question 10:

moderate

The earth (mass $= 6 \times 10^{24}\text{ kg}$) revolves around the sun with an angular velocity of $2 \times 10^{-7}\text{ rad/s}$ in a circular orbit of radius $1.5 \times 10^8\text{ km}$. The force exerted by the sun on the earth, in newton, is:

(1995)

The force is the centripetal force $F = mR\omega^2$. Substituting the values: $F = (6 \times 10^{24}) \times (1.5 \times 10^{11}\text{ m}) \times (2 \times 10^{-7})^2 = 9 \times 10^{35} \times 4 \times 10^{-14} = 36 \times 10^{21}\text{ N}$.

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