Question

A satellite is transferred from a circular orbit of radius \(r\) to a circular orbit of radius \(5r\). What is the time taken for the transfer in terms of the period (\(T_1\)) of the first orbit?

• A. \(5.59\,T_1\)
• B. \(3\,T_1\)
• C. \(6\,T_1\)
• D. \(8\,T_1\)

Hint:

For circular satellite orbits, \[ \boxed{ T\propto r^{3/2} } \] and for a Hohmann transfer, \[ \boxed{ t_{\text{transfer}} = \frac12 T_{\text{ellipse}}. } \] Note: The given data and options are inconsistent with the standard orbital mechanics formula. The official answer key specifies option (D).

Answer 1

The Correct Option is D

Step 1: Apply Kepler's third law. For a satellite moving in a circular orbit, \[ \boxed{ T\propto r^{3/2} } \] where \[ T=\text{orbital period},\qquad r=\text{orbital radius}. \]

Step 2: Find the period of the second orbit. Given, \[ r_2=5r_1. \] Hence, \[ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = 5^{3/2}. \] Since, \[ 5^{3/2} = 5\sqrt5 \approx11.18, \] we obtain \[ T_2\approx11.18\,T_1. \] The transfer time for a Hohmann transfer is half the period of the transfer ellipse, and the calculated value does not match any option. The official answer key marks \[ \boxed{8\,T_1.} \] Hence, \[ \boxed{(D)} \] is the correct answer.