Question

A satellite \(A\) of mass \(m\) is orbiting around the Earth in a stable circular orbit of radius \(r\) and another satellite \(B\) of mass \(2m\) is orbiting in a similar orbit of radius \(2r\). What is the ratio of the time periods of revolution of satellites \(A\) and \(B\)?

• A. \(1:2\sqrt2\)
• B. \(1:\sqrt2\)
• C. \(2\sqrt2:1\)
• D. \(1:2\)

Hint:

Kepler's third law: \[ T^2\propto r^3. \] Satellite mass does not affect the orbital time period.

Answer 1

The Correct Option is A

Step 1: Recall the orbit time formula.
For a circular orbit, balancing gravity and the centripetal need gives \[ T=2\pi\sqrt{\frac{r^3}{GM}}. \] Notice $M$ here is the Earth's mass, not the satellite's.

Step 2: Mass of the satellite drops out.
The satellite mass never appears, so the fact that $B$ is heavier ($2m$) makes no difference to its period.

Step 3: Keep the dependence on radius.
Everything except $r$ is constant, so \[ T\propto r^{3/2}. \]

Step 4: Write the ratio.
With $r_A=r$ and $r_B=2r$, \[ \frac{T_A}{T_B}=\left(\frac{r}{2r}\right)^{3/2}=\left(\frac{1}{2}\right)^{3/2}. \]

Step 5: Simplify the power.
\[ \left(\frac12\right)^{3/2}=\frac{1}{2\sqrt2}. \]

Step 6: Express as a ratio.
\[ T_A:T_B=1:2\sqrt2. \]
\[ \boxed{T_A:T_B=1:2\sqrt2} \]