# An astronaut in space is in orbit 250 km above the surface of Earth. What is the gravitational acceleration on the astronaut at this altitude?

Jun 7, 2016

The acceleration due to gravity is given by $a = \frac{G M}{r} ^ 2$ where G is the gravitational constant and M is the mass of Earth. The value at this altitude is $9.08$ $m {s}^{-} 2$.

#### Explanation:

Constants:

$G = 6.67408 \times {10}^{-} 11$ ${m}^{3} k {g}^{-} 1 {s}^{-} 2$
$M = 5.972 \times {10}^{24}$ $k g$

The radius of Earth is $r = 6.371 \times {10}^{6}$ $m$

In this case, we need to add another $2.5 \times {10}^{5}$ $m$ (250 km) to the radius, to give $r = 6.621 \times {10}^{6}$ $m$

So:

$a = \frac{G M}{r} ^ 2 = \frac{6.67 \times {10}^{-} 11 \times 5.97 \times {10}^{24}}{6.621 \times {10}^{6}} ^ 2 = 9.08$ $m {s}^{-} 2$

This makes sense, since the acceleration due to gravity is $9.81$ $m {s}^{-} 2$. The radius has been increased by only about 4%, but the acceleration due to gravity is inversely proportional to the square of the radius.