Question

$hii \: nirman \: sir$
Explain the Kepler's third law.​

Answer 1

KEPLER'S 3RD LAW:

Statement:

• Kepler's 3rd Law states that the square (2nd power) of the orbital period is directly proportional to the cube (3rd power) of the semi-major axis of the orbit of the planet.

Explanation:

• We know that planets revolve around the sun in an elliptical orbit.

• Now, the elliptical trajectory has a major and a minor axis.

• If we consider a semi-major axis of any planet , then the below equation will be followed:

$\boxed{ \bf{ \large{ {T}^{2} \: \propto \: {R}^{3} }}}$

Here T is time period (orbital period) and R is the semi-major axis length.

• In the statement, orbital period actually represents the total time taken by the planet to complete one revolution around the sun.

$\bf{ \implies \dfrac{ {T}^{2} }{ {R}^{3} } = constant}$

• So, we can say that this ratio of T² and R³ will be constant for all planets.

Kindly refer to the attached diagram to have a proper understanding of the trajectory of a planet around the sun.

Hope It Helps.

Answer 2

$\star$ Kepler's Third Law:-

• It gives a relationship between the time period/orbital period of a satellite and the length of the semi major axis of the orbit.
• It says that the square of the time period/orbital period (T) of a satellite is directly proportional to the cube of the length of the semi major axis of the orbit.
• It can be written as $\boxed{\sf{T^2 \propto r^3}}$
• And hence $\boxed{\sf{\dfrac{T^2}{r^3} = constant}}$

For Derivation of this law, refer to the 1st attachment.

In the derivation, $\sf F_c$ stands for Centripetal force = Mass * Centripetal Acceleration $\sf = m * \dfrac{v^2}{r} = \dfrac{mv^2}{r}$

Now, let's do a numerical using this law!

Q) A satellite is orbiting Earth at a height of 6R above the surface of Earth where R is the radius of Earth. Time period of that satellite is 24 hours. The time period of another satellite at a distance of 3.5R form the centre of the Earth is?

Ans)

⚠️Careful, for first satellite, 6R is the distance between the satellite and the surface of the Earth, while for the second satellite, 3.5R is the distance between the satellite and the centre of the Earth.

Refer to 2nd attachment for the diagram of this answer.

In this questions, we must consider r as the distance between satellite and the centre of the Earth.

Hence, for first satellite, the distance between the satellite and the centre of the Earth will be 7R.

Now, by applying Kepler's Third Law,

$\sf T^2 \propto r^3$

$\longrightarrow \sf{ \bigg( \dfrac{T_1}{T_2} \bigg)^2 = \bigg( \dfrac{R_1}{R_2} \bigg)^3}$

Let $\sf{ T_2}$ be x

$\longrightarrow \sf \bigg( \dfrac{24}{x} \bigg)^2 = \bigg( \dfrac{7 \cancel{R}}{3.5 \cancel{R}} \bigg)^3$

After solving, we will get,

$\longrightarrow \sf T_2 = x = 8.48\: hours$