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Escape Velocity and planet’s gravitational field

  • Thread starter RoryP
  • Start date
  • #1
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Homework Statement


The escape velocity is the velocity of an object at the surface of a planet that would
allow it to be removed completely from the planet’s gravitational field.
Calculate the escape velocity for an object on the surface of Pluto.

mass of Pluto: 1.3 *1022
radius of Pluto: 1.2 *106
G= 6.7 *10-11

Homework Equations


EK= 1/2mv2
Potential Energy= GM1M2/R


The Attempt at a Solution


I think i have this correct but i dont have the answers to check!!! So i need a professional opinion =]
So at the surface would i be right in saying the kinetic energy required to leave the planet will be equal to the Potential the body has at the surface of the planet??
So, 1/2mv2= GM1M2/R
(the mass of the object will cancel)
v=[square root]{2GM/R}
v=[square root]{2*6.7 *10-11*1.3 *1022/1.2 *106
v= 1205 ms-1
Is this the right answer?? not entirely sure!
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4

Homework Statement


The escape velocity is the velocity of an object at the surface of a planet that would
allow it to be removed completely from the planet’s gravitational field.
Calculate the escape velocity for an object on the surface of Pluto.

mass of Pluto: 1.3 *1022
radius of Pluto: 1.2 *106
G= 6.7 *10-11

Homework Equations


EK= 1/2mv2
Potential Energy= GM1M2/R


The Attempt at a Solution


I think i have this correct but i dont have the answers to check!!! So i need a professional opinion =]
So at the surface would i be right in saying the kinetic energy required to leave the planet will be equal to the Potential the body has at the surface of the planet??
So, 1/2mv2= GM1M2/R
(the mass of the object will cancel)
v=[square root]{2GM/R}
v=[square root]{2*6.7 *10-11*1.3 *1022/1.2 *106
v= 1205 ms-1
Is this the right answer?? not entirely sure!
Yes. You need to overcome the potential at the surface. 1/2mV2 = Gmp/rp

Where rp and mp are the radius and mass of the dwarf planet Pluto
 
  • #3
76
0
ahhh sweet cheers!! =]
 
  • #4
174
0
Hi ,
I suggest in question like this in a test:
Don't write right away GMm/R=Mv^2/2
say that Potential energy = -GMm/R notice the negative sign.
and in infinity the energy==0
than Initial kinetic energy + (which will get us minus here) potential energy=0(getting to infinity).

Just An advice from some 1 that suffered this thing on his flesh :D
 

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