Escape Velocity (from surface of the Earth)

AI Thread Summary
The escape velocity for a satellite on Earth's surface is calculated using the formula Vesc = √(2GMp/r). After substituting the values for gravitational constant G, Earth's mass Mp, and Earth's radius r, the calculated escape velocity is approximately 11183 m/s. The poster expresses concern that this value seems too high, questioning their calculations. However, it is confirmed that the escape velocity for Earth is indeed in the range of 11 km/s, validating the calculation. The discussion emphasizes the importance of verifying results against known values.
Tyyoung
Messages
7
Reaction score
0

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.
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top