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Rufina [12.5K]
4 years ago
10

Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4000 km from the surface of

the earth (The mass of the earth is 5.97×1024 kg and the radius of the earth is 6.38×106 m).
Physics
1 answer:
evablogger [386]4 years ago
8 0

v^2 =GM/r^2

the speed of satellite is 3.12*10^6

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height o
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Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy E_{o} must be equal to the final energy  E_{f}:

E_{o}=E_{f} (1)

Where each energy is the sum of kinetic energy K and potential energy U:

K_{o}+U_{o}=K_{f}+U_{f} (2)

Where:

K_{o}=\frac{1}{2}mV_{o}^{2}

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U_{o}=mgh_{o}

Being  g=9.8 m/s^{2} the acceleration due gravity and  h_{o}=25 m your initial height

K_{f}=\frac{1}{2}mV_{f}^{2}

Being V_{f} your final velocity

U_{f}=mgh_{f}

Being h_{f}=3 m your final height

Rewritting (2):

\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f} (3)

mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f}) (4)

Isolating V_{f}:

V_{f}=\sqrt{2g(h_{o}-h_{f})} (5)

V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)} (6)

Finally:

V_{f}=20.765 m/s This is your spedd when you arrive at 3 m height

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3 years ago
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