Escape Velocity (from surface of the Earth)

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SUMMARY

The escape velocity for a satellite on the surface of the Earth is calculated using the formula Vesc = √(2GMp/r), where G is the gravitational constant (6.67x10^-11), Mp is the mass of the Earth (5.98x10^24 kg), and r is the radius of the Earth (6.38x10^6 m). The calculated escape velocity is approximately 11182.966 m/s, confirming that the satellite must exceed this speed to escape Earth's gravitational pull. The discussion clarifies that this value is indeed in the expected range for Earth's escape velocity.

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Homework Statement



A satellite with a mass of 5.00x10^2 kg is in a circular orbit, whose radius is 2re, around Earth. Then it is moved to a circular orbit with a radius of 3re.

e) Calculate the escape velocity for the satellite if it is on Earths surface.

Homework Equations



Vesc = Sq.root of 2GMp/r

The Attempt at a Solution



Vesc = Sq.root of 2(6.67x10^-11)(5.98x10^24)/6.38x10^6

Vesc = Sq.root of 125036363.6
Vesc = 11182.966 m/s
Therefore the velocity of the satellite must be > 11182.966 m/s

I must of done something wrong because 11000m/s seems way to fast! Plz help.
Thanks in advance.
 
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I think that the escape velocity of the Earth in the km/s range. So that should be correct.
 
Use google to check ;] your answer is correct
 

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