Kepler’s Laws Of Planetary Motion MCQ - Practice Questions with Answers

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 planets are large natural bodies rotating around a star in definite orbits.

So, Kepler's 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.

The below figure

Area of velocity $=\frac{dA}{dt}$

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

Where

$$\begin{array}{rl}& \frac{dA}{dt}{\to }_{\text{Areal velocity }}\\ & dA\to \text{ small area traced }\end{array}$$

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

$$\text{ As }\frac{dA}{dt}=\frac{L}{2m}$$

where
$L=mvr\to$ 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.

The below figure

$$\begin{array}{rl}& AB=AF+FB\\ & 2a=r_1+r_2\\ & \therefore a=\frac{r_1+r_1}{2}\end{array}$$

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 $\mathrm{T}=$ Time period of revolution
Then according to Kepler's 3rd law.

$$\begin{array}{rl}& T^2\alpha a^3\\ & T^2\alpha {\left(\frac{r_1+r_2}{2}\right)}^3\end{array}$$