gravity launch velocity

[chiddler]

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If the escape velocity of a rocket from the surface of the Earth is ve, then the escape velocity of the same rocket from the surface of a planet whose acceleration due to gravity as well as radius are 4 times that of the Earth is

A 4ve
B ve/4
C ve
D 16ve

As long as they are moving up, then they should be able to escape the planet, right? So it may take more force to get this same velocity, but it is indeed the same velocity.

"The escape velocity of an object from the surface of a planet of radius R and acceleration due to gravity g is given by (2gR)1/2. Thus, increasing both g and R by four will lead to an escape velocity 4 times higher."

Where did they get that equation from?

answer is

a

 

[SaintJude]

PE=KE
mgh= 1/2mv^2

sqrt (2gh) = v

But I'm almost positive this would be a passage-based question on the MCAT, and you'd be able to derive the equation from a passage given.

 

[chiddler]

PE=KE
mgh= 1/2mv^2

sqrt (2gh) = v

good point.

why is more velocity required, anyway? as long as they're moving up and there is no wind or friction.

 

[MedPR]

Whether or not this is right, its what I did.

F=ma
a=v/t

a=g
4g=4v/t
4v.

 

[milski]

good point.

why is more velocity required, anyway? as long as they're moving up and there is no wind or friction.

Launch velocity for a rocket is when the engines stop and the rocket continues with whatever velocity it has at that time. If it's not fast enough, it will fall back to the Earth or stay in orbit around it.

 

[chiddler]

Launch velocity for a rocket is when the engines stop and the rocket continues with whatever velocity it has at that time. If it's not fast enough, it will fall back to the Earth or stay in orbit around it.

ah this makes more sense.

Then the kinetic energy must be sufficient to overcome the gravitational energy.

thus the 1/2 mv^2 = mgh