A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity (3π/G) is equal to

Question

A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity (3π/G) is equal to (A) T2D (B) 3π T2D (C) 3π D2T (D) D2T

Tardigrade Admin · Accepted Answer

Using T = 2 π √(R3/GM) = 2π √(R3/G × (4)3 × π R3 D) T2 = (4 π2 R3/G (4)3 π R D) = (3 π/DG) (3 π/DG) = T2 D

Q. A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity $\frac{3 π}{G}$ is equal to

$T^{2} D$

$3 π T^{2} D$

$3 π D^{2} T$

$D^{2} T$

Using $T = 2 π \frac{R ^{3}}{GM} = 2 π \frac{R ^{3}}{G \times \frac{4}{3} \times π R ^{3} D}$
$T^{2} = \frac{4 π ^{2} R ^{3}}{G \frac{4}{3} π R D} = \frac{3 π}{D G}$
$\frac{3 π}{D G} = T^{2} D$

Q. A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity G3π​ is equal to

T2D

3πT2D

3πD2T

D2T