Question

The gravitational constant may also be calculated by analyzing the motion of a rocket. Suppose a rocket is launched vertically from the surface or Earth at an initial speed of vi. Its initial distance from the center of Earth is Ri, the radius of Earth. Its peak distance, where its speed is momentarily zero is, is Rf. For simplicity, ignore air resistance and Earth’s rotation. Enter an expression for the gravitational constant, in terms of vi, Ri, Rf, and ME.

Answer 1

Answer:

$G = \frac{v_i^2}{2M_E(\frac{1}{R_i} - \frac{1}{R_f})}$

Explanation:

As we know that during the projection of rocket the total mechanical energy is conserved as there is no friction in it's motion

So we will have

$\frac{1}{2}mv_i^2 - \frac{GM_Em}{R_i} = 0 - \frac{GM_Em}{R_f}$

now divide whole equation by mass of the object

$v_i^2 - \frac{2GM_E}{R_i} = - \frac{2GM_E}{R_f}$

$v_i^2 = 2GM_E(\frac{1}{R_i} - \frac{1}{R_f})$

now we can rearrange above equation to find gravitational constant

$G = \frac{v_i^2}{2M_E(\frac{1}{R_i} - \frac{1}{R_f})}$

So above is the expression for gravitational constant G