Question

The value of g at the height of the space shuttle’s orbit is

Answer 1

Answer:

$g'=9.3345\ m.s^{-2}$

Explanation:

The minimum height of a space shuttle from the surface of the earth is 304 kilometers.

So according to this data we calculate the value to g at that height.

We have the formula:

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

where:

g' = the acceleration due to gravity at the given height 'h' above the earth

R = radius of the earth

g = the acceleration due to gravity at the surface of the earth

Now, using eq. (1)

$g'=9.8(1-\frac{304\times 10^3}{64000\times 10^3} )$

$g'=9.3345\ m.s^{-2}$