# The Apollo spacecraft must be traveling at what speed in order to remain in a 110 km orbit around the moon?

Question by Armani: The Apollo spacecraft must be traveling at what speed in order to remain in a 110 km orbit around the moon?
Long ago, Sir Isaac Newton gave us a mathematical description of how one object affects, and is affected by, the gravitational force of another object. Many, many years of observations have proven this description to be accurate (at least for masses like those of the planets). Newton’s Law of Gravitation states: The force between any two objects having masses M1 and M2 separated by a distance R is an attraction along the line joining the objects and has a magnitude of:

F = (G x M1 x M2) / (R x R).

G is the universal gravitational constant, which has a value of 6.6732 x 10-11 newton-meters2/kg2 for all pairs of objects. (A “newton” is a unit of force that physicists use. It is defined to be the amount of force needed to accelerate a 1 kg mass at 1 meter per sec2. A newton, as a unit of force, is fairly small, like a millimeter is a small unit of distance or a microsecond is a small unit of time.)
How do we know what speed an object in orbit around a planet or a moon must travel to maintain it’s orbit and not be pulled down to the surface by gravity? The orbital velocity equation tells us how!

Question 1: The Apollo spacecraft must be traveling at what speed in order to remain in a 110 kilometer orbit around the moon?

The magnitude of the velocity can be computed exactly from the laws of gravitational motion. To remain in orbit, a spacecraft must travel at a very high velocity. The required velocity is dependent on gravity and decreases with increasing altitude (i.e., distance) as shown:

v=(GM/r)1/2 or v= SQRT (GM/r)

where V is the orbital velocity, R is the radius of the orbit, and G is the local acceleration of gravity. You can work the problem from scratch or use the shortcut below:

Shortcut: GM = 0.0049 (106 kilometers3/seconds2)

Hint: First find the radius of the moon and then add that to the orbital altitude to answer the problem!

Question 2: Calculate the escape velocity of the moon.
The escape velocity (vesc) of a body depends on the mass (M) and the radius (r) of the given body. The formula which relates these quantities is:

vesc = (2 * G * M / r)1/2

where G is called the Gravitational constant.

The notation

(2 * G * M / r)1/2

means (2 * G * M / r) to the one-half power, which is equal to the square root of (2 * G * M / r).

You will calculate the escape velocity for the Moon using the MKS system where the unit for distance is meters, the unit for mass is kilograms, and the unit for time is seconds.

In this system, the gravitational constant has the value:

G = 6.67 * 10-11 meter3/kilogram-seconds2.

As an example, the mass M of the Earth is 5.98 * 1024 kilograms. The radius r of the Earth is 6378 kilometers, which is equal to 6.378 * 106 meters. The escape velocity at the surface of the Earth can therefore be calculated by:

vesc
=
(2 * G * M / r)1/2

=
( 2 * (6.67 * 10-11) * (5.98 * 1024) / (6.378 * 106) ) 1/2

=
1.12 * 104 meters/second

=
11.2 kilometers/second

This simple physics equation can be used to calculate the escape velocity for any body if you know the mass of the body and its radius!

The escape velocity for the Earth is therefore 11.2 kilometers per second. This is the velocity that an object needs at the surface of the Earth to be able to overcome the gravitational attraction of the Earth and escape to space.

Use the Internet to find the mass and radius of the Moon. Make sure to convert the radii from kilometers to meters when making the calculation, and make sure that you can calculate the escape velocity correctly.