Question

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

Answer 1

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

Step-by-step 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:

`a_(r) = \omega^(2)\cdot R` (Eq. 1)

Where:

`\omega` - Angular velocity, measured in radians per second.

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

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

`a_(r) = G\cdot (M)/(R^(2))` (Eq. 2)

Where:

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

`M` - Mass of the Earth, measured in kilograms.

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

`\omega^(2)\cdot R = G\cdot (M)/(R^(2))`

And we solve for
`R`:

`R^(3) = (G\cdot M)/(\omega^(2))`

`R = \sqrt[3]{(G\cdot M)/(\omega^(2)) }` (Eq. 3)

But we know also that angular velocity is:

`\omega = (2\pi)/(T)` (Eq. 4)

Where
`T` is the period of rotation, measured in seconds.

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

`R = \sqrt[3]{(G\cdot M\cdot T^(2))/(4\pi^(2)) }` (Eq. 5)

If we know that
`G = 6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2))`,
`M = 5.98* 10^(24)\,kg` and
`T = 107928\,s`, then the distance of the satellite from the center of the Earth is:

`R = \sqrt[3]{(\left(6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)) \right)\cdot (5.98* 10^(24)\,kg)\cdot (107928\,s)^(2))/(4\pi^(2)) }`

`R \approx 49.015* 10^(6)\,m`

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

`R \approx 30463.228\,mi`

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

`d = 30463.228\,mi-3958.8\,mi`

`d = 26504.428\,mi`

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