Question

A starship is circling a distant planet of radius R. The astronauts find that the free-fall acceleration at their altitude is half the value at the planet's surface. How far above the surface are they orbiting?

Answer 1

Answer:

h = R/4 = 1.5925 x 10⁶ m = 1592.5 km

Explanation:

The variation in the value of acceleration due to gravity value with respect to the altitude is given by the following general formula:

$g' = g(1-2\frac{h}{R})$

where,

h = altitude from surface of earth = ?

g = acceleration due to gravity on surface of earth = 9.81 m/s²

R = Radius of Earth = 6.37 x 10⁶ m

g' = acceleration due to gravity at given altitude = g/2

Therefore,

$\frac{g}{2} = g(1-2\frac{h}{R})\\\\2\frac{h}{R} = 1-\frac{1}{2}\\\\2\frac{h}{R} = \frac{1}{2}\\\\\frac{h}{R} = \frac{1}{4}$

h = R/4 = 1.5925 x 10⁶ m = 1592.5 km