Kepler's third law states that… - NEET UG 2015 Physics PYQ

NEET UG 2015 — Physics Mechanics

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Kepler's third law states that the square of the period of revolution $\text{(T)}$ of a planet around the sun is proportional to the third power of the average distance $\text{r}$ between sun and planet i.e., ${T}^{2}=K{r}^{3}$ .

Here $\text{K}$ is constant. If the masses of sun and planet are $\text{M}$ and $\text{m}$ respectively then as per Newton's law of gravitation force of attraction between them is

$F=\frac{GMm}{{r}^{2}}$ , here $\text{G}$ is gravitational constant.

The relation between $\text{G}$ and $\text{K}$ is described as:

Official previous-year question

Held on 30 Apr 2015 · Verified 9 Jul 2026.

Options

A
$GK=4{\pi }^{2}$
B
$GMK=4{\pi }^{2}$
C
$K=G$
D
$K=\frac{1}{G}$

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Solution

Here gravitational force is equal to centripetal force required for motion of planet.

Hence,

$\frac{GMm}{{r}^{2}}=\frac{m{v}^{2}}{r}$

${v}^{2}=\frac{GM}{r}$

Time period is given by,

$T=\frac{2\pi r}{v}$

${T}^{2}=\frac{4{\pi }^{2}{r}^{2}}{{v}^{2}}$

${T}^{2}=\frac{4{\pi }^{2}{r}^{2}}{(\frac{Gm}{r})}=\frac{4{\pi }^{2}{r}^{3}}{GM}$

We have given,

${T}^{2}=k{r}^{3}$

On comparing,

$k=\frac{4{\pi }^{2}}{GM}\Rightarrow GMk=4{\pi }^{2}$

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Solution

Here gravitational force is equal to centripetal force required for motion of planet.

Hence,

$\frac{GMm}{{r}^{2}}=\frac{m{v}^{2}}{r}$

${v}^{2}=\frac{GM}{r}$

Time period is given by,

$T=\frac{2\pi r}{v}$

${T}^{2}=\frac{4{\pi }^{2}{r}^{2}}{{v}^{2}}$

${T}^{2}=\frac{4{\pi }^{2}{r}^{2}}{(\frac{Gm}{r})}=\frac{4{\pi }^{2}{r}^{3}}{GM}$

We have given,

${T}^{2}=k{r}^{3}$

On comparing,

$k=\frac{4{\pi }^{2}}{GM}\Rightarrow GMk=4{\pi }^{2}$

Options

Did you get this right?

Solution

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Options

Did you get this right?

Solution

Your note

NEET UG Mechanics in other years

More NEET UG Mechanics questions

Other Physics topics for NEET UG

Electromagnetism

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Mechanics