Calculate g for space-shuttle trajectory, 200 km above the Earth’s surface. How close is this to 9.8 m/s2 ?

Question

Earth’s mass is 6 x 10^24 kg and it radius is 6.38 x 10^6 m.

Answer 1

To calculate the value for the acceleration due to gravity...also known as the gravitational field strength, g for an object away from the surface of a planet we use the following:

g = GM_planet/R_orbit²

In the current problem the space shuttle is located a distance of 200 km above the surface of the Earth. The orbital radius must take into account both the radius of the Earth and the altitude of the space shuttle. Taking this into account, we know the following values:

M_planet = 6 X 10^24 kg

R_orbit = R_planet + Altitude = 6.38 X 10^6 + 200X10^3 = 6.58 X 10^6 m

G = 6.67 X 10^-11 Nm^2/kg^2

Plugging in our values we get the following:

g = Gm_planet/R_orbit² = (6.67X10^-11)(6X10^24)/(6.58X10^6)^2

g = 9.24 m/s^2

You will note that this value is off of the accepted value for gravity on the surface of earth (9.8 m/s^2) by only ~0.556 m/s^2. This is because the altitude is relatively small compared to the radius of the earth and as such the change is not great.