Question

A satellite P takes 2 years to make a circular orbit of radius R around the earth. The time period of another satellite Q if it moves in a circular orbit twice the orbital radius of satellite P around earth is:

• A. 5.2 years
• B. 5.1 years
• C. 5.6 years
• D. 5.9 years

Hint:

For satellites, time period increases as $r^{3/2}$, so even small increases in radius significantly increase orbital time.

Answer 1

The Correct Option is C

Step 1: Derive the period-radius relation from Newton's law of gravitation, instead of quoting Kepler's third law directly.
For a satellite of mass $m$ moving in a circular orbit of radius $r$ around a planet of mass $M_e$, gravity supplies the centripetal force: \[ \frac{GM_em}{r^2} = m\omega^2 r = m\left(\frac{2\pi}{T}\right)^2 r \] Solving for $T^2$: \[ T^2 = \frac{4\pi^2 r^3}{GM_e} \] So $T^2 \propto r^3$ follows directly from force balance, with the same constant of proportionality for both satellites since they orbit the same planet.
Step 2: Write the ratio for the two satellites.
\[ \left(\frac{T_Q}{T_P}\right)^2 = \left(\frac{r_Q}{r_P}\right)^3 \]
Step 3: Substitute the given radius ratio.
Satellite Q orbits at twice the radius of P, so $r_Q/r_P = 2$: \[ \left(\frac{T_Q}{T_P}\right)^2 = 2^3 = 8 \]
Step 4: Take the square root.
\[ \frac{T_Q}{T_P} = \sqrt{8} = 2\sqrt{2} \approx 2.828 \]
Step 5: Substitute $T_P = 2$ years.
\[ T_Q = 2 \times 2.828 = 5.656 \ \text{years} \approx 5.6 \ \text{years} \]
Final Answer:
\[ \boxed{5.6 \ \text{years}} \]