NEET Keplers Law MCQs & PYQs

Practice NEET Keplers Law 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 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 3:

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 4:

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 5:

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 6:

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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