Question

A satellite is put into a circular orbit very close to the surface of the Earth, just outside the atmosphere. What is its orbital speed, and how is this speed related to the acceleration due to gravity and the radius of the Earth?

Hint:

Set the gravitational force equal to the centripetal force required for circular motion, then use \( g = GM/R^2 \) to remove G and M from the expression.

Answer 1

Step 1: A satellite moving in a circular orbit close to the Earth's surface is held in that path by the gravitational pull of the Earth. This gravitational force supplies the centripetal force needed for circular motion.

Step 2: Equating the gravitational force to the centripetal force for a satellite of mass m at a distance equal to the Earth's radius R from the centre:
\( \dfrac{GMm}{R^2} = \dfrac{mv^2}{R} \)

Step 3: Cancelling m and one factor of R from both sides gives:
\( v^2 = \dfrac{GM}{R} \)

Step 4: The acceleration due to gravity at the Earth's surface is defined as \( g = \dfrac{GM}{R^2} \), so \( GM = gR^2 \). Substituting this into the expression for v squared:
\( v^2 = \dfrac{gR^2}{R} = gR \)

Step 5: Taking the square root gives the orbital speed:
\( v = \sqrt{gR} \)

Step 6: Putting in the standard values \( g = 9.8 \ m/s^2 \) and \( R = 6.4 \times 10^6 \ m \):
\( v = \sqrt{9.8 \times 6.4 \times 10^6} \approx 7.9 \times 10^3 \ m/s \), that is about 7.9 kilometre per second. This is also called the first cosmic velocity, the minimum speed needed to keep a body moving in a circular orbit just above the Earth's surface.