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# Kepler’s Laws Of Planetary Motion - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Kepler’s Laws of Planetary Motion is considered one of the most asked concept.

• 18 Questions around this concept.

## Solve by difficulty

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is    the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis.  If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :

## Concepts Covered - 1

Kepler’s Laws of Planetary Motion

Kepler gives three empirical laws which govern the motion of the planets which are known as Kepler’s laws of planetary motion.

As we know that planets are large natural bodies rotating around a star in definite orbits.

So, Kepler laws are-

(a) The law of Orbits:

It is Kepler's First Law.

Every planet moves around the sun in an elliptical orbit. And the sun will be at one of the foci of the ellipse.

(b) The law of Area:

It is Kepler's 2nd law.

According to this, the line joining the sun to the planet sweeps out equal areas in equal intervals of time which clearly means that areal velocity is constant. So according to this law, a planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. You can find it similar to the law of conservation of angular momentum.

For the below figure

Area of velocity = $\frac{dA}{dt}$

$\frac{dA}{dt}=\frac{1}{2}\frac{\left ( r \right )\left ( Vdt \right )}{dt}=\frac{1}{2}rV$

Where

$\frac{dA}{dt}\rightarrow$ Areal velocity

$dA\rightarrow$  small area traced

Kepler's 2nd law is Similar to the Law of conservation of momentum

As    $\frac{dA}{dt}= \frac{L}{2m}$

where

$L=mvr\rightarrow$ Angular momentum

(c) The law of periods:

It is Kepler's 3rd law.

According to this, the square of the Time period of revolutions of any planet around the sun is directly proportional to

the cube of the semi-major axis of that particular orbit.

For the below figure

$AB=AF+FB$

$2a=r_{1}+r_{2}$

$\therefore\; a=\frac{r_{1}+r_{1}}{2}$

Where

$a=$ semi major Axis

$r_{1}=$ The shortest distance of the planet from the sun (perigee)

$r_{2}=$  Largest distance of the planet from the sun (apogee)

So if T=Time period of revolution

Then according to Kepler's 3rd law.

$T^{2}\: \alpha\: a^{3}$

or  $T^{2}\: \alpha \: \left ( \frac{r_{1}+r_{2}}{2} \right )^{3}$

## Study it with Videos

Kepler’s Laws of Planetary Motion

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