Relating escape velocity equations

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SUMMARY

The discussion centers on the relationship between the escape velocity equations, specifically sqrt(2GM/r) and sqrt(2gr). Participants clarify that both equations represent the same physical concept, with the gravitational force equations being central to the derivation. The key step involves equating the gravitational force F = -Gm1m2/r^2 with the force near the Earth's surface, F = mg, leading to the simplification that connects the two escape velocity equations. The final insight is substituting g = Gm/R^2 into sqrt(2gr) to establish the equivalence.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = -Gm1m2/r^2
  • Familiarity with escape velocity concepts and equations
  • Basic algebra skills for manipulating equations
  • Knowledge of gravitational acceleration near Earth's surface, g = Gm/R^2
NEXT STEPS
  • Study the derivation of escape velocity from gravitational force equations
  • Learn about gravitational potential energy and its relation to escape velocity
  • Explore the implications of escape velocity in astrophysics and space travel
  • Investigate the differences between escape velocity and orbital velocity
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and gravitational theory, as well as educators seeking to clarify concepts related to escape velocity and gravitational forces.

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
 
Last edited:
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Do you know the equation for the gravitational force between two objects (in space)?
 
F=-Gm1m2/r^2
 
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 F = -Gm_1m_2/r^2. 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.
 
Last edited:
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:
 
Just what I was going to say :wink: thanks for jumping in on that, tiny-tim.
 

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