Math Exploration 7.2: Gravitational Acceleration between Two Astronomical Objects

In this example, we will calculate the gravitational acceleration in the Earth-Moon system.

If Earth exerts a force on the Moon, causing it to move in a near-circular orbit around Earth, then the Moon must exert an equal and opposite force on Earth, causing it to accelerate toward the Moon. This is a balanced action-reaction pair, so no net force acts on the Earth-Moon system as a whole.

However, these forces do accelerate each object separately, and we can find the magnitude of Earth’s acceleration, aE, using a combination of Newton’s law of gravitation and Newton’s second law:

aE = FE / ME = GMMME/ME r2M = GMM / r2M

Here the force on Earth from the Moon is FE, ME is the mass of Earth, MM is the Moon’s mass, and rM is the distance separating Earth from the Moon. Numerically we get:

aE = ( 6.67384e-11 N m2 kg-2 ) ( 7.3459e22 kg ) / ( 3.844e8 m )2 = 3.4675 x 10-5 m/s2

Calculating the acceleration of the Moon toward Earth we find:

aM = GME / r2M = ( 6.67384e-11 N m2 kg-2 ) (5.98e24 kg ) / ( 3.844e8 m )2 = 2.7e-3 m/s2

Since Earth is about 80 times more massive than the Moon, equal forces exerted on each results in a higher acceleration of the Moon by a factor of ~80.