Modeled after the two satellites problem (slide 12 in Lecture13, Universal gravity), imagine now if the period of the satellite is exact 29.980 hours, the earth mass is 5.98 x 1024 kg, and the radius of the earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in MILES

The distance of the satellite from the surface of the Earth is 26504.428 miles.

Explanation:

From Rotation Physics we remember that satellites moves in a uniform circular motion, in which radial acceleration experimented by the satellite is determined by:

(Eq. 1)

Where:

- Angular velocity, measured in radians per second.

- Distance of the satellite from center of the Earth, measured in meters.

Besides, the radial acceleration is the gravitational acceleration, whose formula is:

(Eq. 2)

Where:

- Gravitational constant, measured in newtons-square meters per square kilogram.

- Mass of the Earth, measured in kilograms.

By equalizing (Eqs. 1, 2), we get the following expression:

And we solve for :

(Eq. 3)

But we know also that angular velocity is:

(Eq. 4)

Where is the period of rotation, measured in seconds.

And we obtain the following expression by applying (Eq. 4) in (Eq. 3):

(Eq. 5)

If we know that , and , then the distance of the satellite from the center of the Earth is:

A mile equals to 1609 meters, then we find that:

Lastly, the distance of the satellite from the surface of the Earth is obtained by subtracting the radius of the planet from previous result:

The distance of the satellite from the surface of the Earth is 26504.428 miles.