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

Answer:

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:

$a_{r} = \omega^{2}\cdot R$ (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:

$a_{r} = G\cdot \frac{M}{R^{2}}$ (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:

$\omega^{2}\cdot R = G\cdot \frac{M}{R^{2}}$

And we solve for $R$:

$R^{3} = \frac{G\cdot M}{\omega^{2}}$

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

But we know also that angular velocity is:

$\omega = \frac{2\pi}{T}$ (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):

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

If we know that $G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$, $M = 5.98\times 10^{24}\,kg$ and $T = 107928\,s$, then the distance of the satellite from the center of the Earth is:

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

$R \approx 49.015\times 10^{6}\,m$

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.