Relating escape velocity equations

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calfly
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Homework Statement


I'm having a hard time relating the escape velocity equations, sqrt(2GM/r) to sqrt(2gr)

The Attempt at a Solution


I know that both equations mean the same thing, but I can't figure out how to get from the first equation to the second
 
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Yep, and I assume you know the equation for the gravitational force acting on an object on (or near) the surface of the Earth... now, the force should be the same whether you calculate it using one equation or the other. So set the two forces equal to each other and see what you get.
 
err, I think this is where I'm stuck at. if an object is right on the surface of the Earth then I guess the gravitational force would look the same as the general equation-ish
F=-Gm(earth)m(object)/R(earth)^2. I'm not quite sure what I'm looking at when I set them equal to each other.
 
calfly said:
F=-Gm(earth)m(object)/R(earth)^2
OK, well, that's right... now, when I mentioned the equation for gravitational force on an object near the surface of the Earth, I was trying to get you to think of the equation you used before you learned about [itex]F = -Gm_1m_2/r^2[/itex]. Do you remember?
 
Ah I think I'm starting to get it but I'm making a math error somewhere i think,
okay so I set f = mg = -Gm1m/r^2, cancel out "m", --> g=-Gm1/r^2.

now I got to fit that into sqrt(2Gm/R) but that does not get sqrt(2gr)?
err i think.
 
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Welcome to PF!

Hi calfly! Welcome to PF! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
calfly said:
Ah I think I'm starting to get it but I'm making a math error somewhere i think,
okay so I set f = mg = -Gm1m/r^2, cancel out "m", --> g=-Gm1/r^2.

now I got to fit that into sqrt(2Gm/R) but that does not get sqrt(2gr)?
err i think.

( diazona :smile: seems to be offline, so …)

Put g = Gm/r2 into √(2gr). :wink: