Question

A satellite is revolving around planet of mass $2M$ in orbit of radius $R$ with time period is $T_1$. Another satellite is revolving around planet of mass $4M$ in orbit of radius $2R$, with time period $T_2$. Find $\frac{T_1}{T_2}$.

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The Correct Option is C

To solve this problem, we will use Kepler's Third Law, which relates the time period of orbit to the radius of the orbit and the mass of the central body. The law is given by:

\(T^2 \propto \frac{R^3}{M}\)

Here, \(T\) is the time period of the orbit, \(R\) is the radius of the orbit, and \(M\) is the mass of the planet around which the satellite orbits.

We have two scenarios to compare:

1. \(T_1\): A satellite revolves around a planet of mass \(2M\) and orbit radius \(R\).
2. \(T_2\): Another satellite revolves around a planet of mass \(4M\) and orbit radius \(2R\).

Applying Kepler's Third Law for each case:

• For the first satellite: \(T_1^2 \propto \frac{R^3}{2M}\)
• For the second satellite: \(T_2^2 \propto \frac{(2R)^3}{4M} = \frac{8R^3}{4M} = \frac{2R^3}{M}\)

Now, set up the ratio using Kepler's Third Law:

\(\left(\frac{T_1}{T_2}\right)^2 = \frac{\frac{R^3}{2M}}{\frac{2R^3}{M}}\)

Simplifying the above expression:

\(\left(\frac{T_1}{T_2}\right)^2 = \frac{R^3 \cdot M}{2M \cdot 2R^3} = \frac{1}{4}\)

By taking the square root of both sides:

\(\frac{T_1}{T_2} = \frac{1}{2}\)

Thus, the correct answer is the ratio \(\frac{T_1}{T_2} = \frac{1}{2}\).

Therefore, the correct option is: \(\frac{1}{2}\).