Question

The International Space Station (ISS) orbits the Earth at a height of about 370 km above the Earth's surface, which equals about 0.058 of Earth's radius. What is g (in m/s2) at the orbit of the ISS?

Answer 1

Answer:

The acceleration due to gravity at the orbit of the ISS is  8.70 m/s²

Explanation:

The magnitude of Earth's gravitational field g at a distance r from the center of Earth is given as;

$g = G\frac{M_{Earth}}{r^2}$

The Space Station is located at a distance r

r = Radius of Earth + altitude of the space Station

r = 6400km + 370km

r = 6770km = 6.77 X 10⁶ m

G is Newton's gravitational constant = 6.6742 X 10⁻¹¹m³/kg⋅s²

mass of Earth = 5.972 X  10²⁴ kg

g = (6.6742 X 10⁻¹¹ * 5.972 X  10²⁴)/(6.77 X 10⁶)²

g = 8.70 m/s²

Answer 2

The acceleration because of gravity at the orbit of the ISS is 8.70 m/s²

Calculation of g:

The Space Station is located at a distance r

So,

r = Radius of Earth + altitude of the space Station

r = 6400km + 370km

r = 6770km = 6.77 X 10⁶ m

Now

G is Newton's gravitational constant = 6.6742 X 10⁻¹¹m³/kg⋅s²

So,

mass of Earth =  5.972 X  10²⁴ kg

g = (6.6742 X 10⁻¹¹ * 5.972 X  10²⁴)/(6.77 X 10⁶)²

g = 8.70 m/s²

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