Question

Correct formula for height of a satellite from earths surface is :

• A. $\left( \frac{T^2 R^2 g}{4 \pi} \right)^{1/2} - R$
• B. $\left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R$
• C. $\left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{1/3} - R$
• D. $\left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{-1/3} + R$

Answer 1

Answer: (B)

$\left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R$

Answer 2

1. Using Gravitational Force and Centripetal Force:
For a satellite of mass m orbiting Earth at height h above the surface, the gravitational force provides the required centripetal force:
GMm / (R + h)² = mv² / (R + h).
Simplifying, we get:
GM / (R + h) = v². (1)

2. Relate Orbital Velocity to Period:
The orbital velocity v can also be expressed in terms of the orbital period T :
v = 2π(R + h) / T. (2)

3. Equate Gravitational Force with Centripetal Acceleration:
We know that GM = gR² (where g is the acceleration due to gravity on Earth's surface). Substituting this in equation (1):
gR² / (R + h) = v².

4. Combine Equations (1) and (2):
Substitute v from equation (2) into the above expression:
gR² / (R + h) = (2π(R + h) / T)².
Rearranging gives:
T²R²g / (2π)² = (R + h)³.

5. Solve for Height _h_:****
Taking the cube root of both sides, we get:
R + h = (T²R²g / 4π²)^(1/3).
Therefore,
h = (T²R²g / 4π²)^(1/3) - R.

Answer:_ (T²R²g / 4π²)^(1/3) - R_(Option 2)