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An artificial satellite circling the Earth completes each orbit in 130 minutes. (a) Find the altitude of the satellite. Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) What is the value of g at the location of this satellite

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Answer:

a) The altitude of the satellite is approximately 2129 kilometers.

b) The gravitational acceleration at the location of the satellite is 5.517 meters per square second.

Step-by-step explanation:

a) At first we assume that Earth is a sphere with a uniform distributed mass and the satellite rotates on a circular orbit at constant speed. From Newton's Law of Gravitation and definition of uniform circular motion, we get the following identity:


G\cdot (m\cdot M)/((R_(E)+h)^(2)) = m\cdot \omega^(2)\cdot (R_(E)+h) (Eq. 1)

Where:


G - Gravitational constant, measured in newton-square meters per square kilograms.


m - Mass of the satellite, measured in kilograms.


M - Mass of the Earth, measured in kilograms.


\omega - Angular speed of the satellite, measured in radians per second.


R_(E) - Radius of the Earth, measured in meters.


h - Height of the satellite above surface, measured in meters.

Then, we simplify the formula and clear the height above the surface:


G\cdot M = \omega^(2)\cdot (R_(E)+h)^(3)


(R_(E)+h)^(3) =(G\cdot M)/(\omega^(2))


R_(E)+h = \sqrt[3]{(G\cdot M)/(\omega^(2)) }


h = \sqrt[3]{(G\cdot M)/(\omega^(2)) }-R_(E)

From Rotation physics, we know that angular speed is equal to:


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

And (Eq. 1) is now expanded:


h = \sqrt[3]{(G\cdot M\cdot T^(2))/(4\pi^(2)) }-R_(E) (Eq. 3)

If we know that
G = 6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)),
M = 5.972* 10^(24)\,kg,
T = 7800\,s and
R_(E) = 6.371* 10^(6)\,m, then the altitude of the satellite is:


h = \sqrt[3]{(\left(6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)) \right)\cdot (5.972* 10^(24)\,kg)\cdot (7800\,s)^(2))/(4\pi^(2)) }-6.371* 10^(6)\,m


h\approx 2.129* 10^(6)\,m


h \approx 2.129* 10^(3)\,km

The altitude of the satellite is approximately 2129 kilometers.

b) The value for the gravitational acceleration of the satellite (
g), measured in meters per square second, is derived from the Newton's Law of Gravitation, that is:


g = G\cdot (M)/((R_(E)+h)^(2)) (Eq. 4)

If we know that
G = 6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)),
M = 5.972* 10^(24)\,kg,
R_(E) = 6.371* 10^(6)\,m and
h\approx 2.129* 10^(6)\,m, then the value of the gravitational acceleration at the location of the satellite is:


g = (\left(6.674* 10^(-11)\,(N\cdot m^(2))/(kg^(2)) \right)\cdot (5.972* 10^(24)\,kg))/((6.371* 10^(6)\,m+2.129* 10^(6)\,m)^(2))


g = 5.517\,(m)/(s^(2))

The gravitational acceleration at the location of the satellite is 5.517 meters per square second.

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