Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

Question

1 year ago

5

s of Earth = 6.37×10^3 km, mass of Earth = 5.98×10^24 kg, G = 6.67×10^-11 Nm^2/kg^2.)

2 answers:

uranmaximum [27] · Answer 1 · 2023-04-11T01:58:30+03:00

In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.

In order to get the solution, we must understand the satellite's planetary motion equation.

<h3>What is the satellite's orbital motion equation?</h3>

The earth's mass M and the satellite's mass M are attracted to one another by gravity.

$F_g=\frac{GMm}{r^2}$

$F_c=\frac{MV^2}{r}$

When a satellite orbits the earth, these two forces are equivalent. Thus, the velocity will be,

$\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s$

As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.

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nignag [31] · Answer 2 · 2023-04-11T00:21:58+03:00

Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.

Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.

<h3>What is the equation of motion of a satellite around earth?</h3>

We have gravitational force of attraction between the satellite of mass m and earth of mass M as,

$F_g=\frac{GMm}{r^2}$

$F_c=\frac{mv^2}{r} \\$

$\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }$

<h3>How to solve the problem?</h3>

$r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2$

$v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s$

Thus, we can conclude that the speed of satellite will be 4188.11 m/s.

Learn more about motion of satellites here:

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