Question 39: A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity \frac{3 \pi}{G d} represents :
(1) T^{3}
(2) \sqrt{T}
(3) T
(4) T^{2}
Answer: Option (4)
Explanation:
For a satellite orbiting just above the surface of the earth, the time period is given by
T = 2\pi \sqrt{\frac{R^{3}}{G M}}.
The mass of the earth can be written in terms of its density as
M = \frac{4}{3}\pi R^{3} d.
Substituting this value of M in the expression for T, we get
T = 2\pi \sqrt{\frac{R^{3}}{G \left(\frac{4}{3}\pi R^{3} d\right)}}.
On simplification,
T = 2\pi \sqrt{\frac{3}{4\pi G d}}.
Squaring both sides,
T^{2} = \frac{4\pi^{2} \times 3}{4\pi G d}.
This gives
T^{2} = \frac{3\pi}{G d}.
Hence, the quantity \frac{3 \pi}{G d} represents T^