Question

A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's centre. Their time periods are in the ratio of

• A. 1:2
• B. 1:16
• C. 1:32
• D. 1:2√2

Hint:

According to Kepler's Law, the time period does not depend upon the mass of the satellite it only depends upon the orbital radius.

Answer 1

The Correct Option is D

The mass of a satellite does not affect the time period 
\(\frac{T_{A}}{T_{B}}=\left(\frac{r_{1}}{r_{1}}\right)^{3 / 2}=\left(\frac{r}{2 r}\right)^{3 / 2}=\left(\frac{1}{8}\right)^{1 / 2}=\frac{1}{2 \sqrt{2}}\)
\(\therefore \) T_A= T_B = 1:2√2

Hence, the correct option is d) 1:2√2.

Answer 2

Correct option (d) \(1:2\sqrt2\)

Explanation: According to Kepler's Law, the time period does not depend upon the mass of the satellite it only depends upon the orbital radius.

As T² ∝ r³

T_A/T = 

\[{r^{3/2} \over 2 ^{3/2} r^{3/2}}\]

So, their time periods are in the ratio of 1:2√2.

Answer 3

The correct option is D)

The mass of the satellite does not affect the time period

\(\frac{T_A}{T_B} = (\frac{r_1}{r_2})\frac{3}{2}\)  

= \((\frac{r}{2r})\frac{3}{2}\)

= \((\frac{1}{8})\frac{1}{2}\)

= \(\frac{1}{2}\sqrt{2}\)

∴ \(T_A =T_B\) = 1:2\(\sqrt{2}\)