Escape Velocity and planet’s gravitational field

In summary, the escape velocity is the minimum velocity an object must have at the surface of a planet to completely leave its gravitational field. To calculate the escape velocity for an object on the surface of Pluto, the equations EK= 1/2mv2 and Potential Energy= GM1M2/R are used. By setting these two equations equal to each other and solving for v, we get the formula v=[square root]{2GM/R}. Plugging in the values for the mass and radius of Pluto, as well as the gravitational constant G, we get an escape velocity of 1205 ms-1. It is important to note that the potential energy at the surface of the planet is negative, and in order to escape,
  • #1
RoryP
75
0

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 don't 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!
 
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  • #2
RoryP said:

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 don't 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
ahhh sweet cheers! =]
 
  • #4
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
 

FAQ: Escape Velocity and planet’s gravitational field

What is escape velocity?

Escape velocity is the minimum speed that an object needs to reach in order to break free from the gravitational pull of a planet and enter into an orbit around it.

How is escape velocity calculated?

The formula for escape velocity is √(2GM/r), where G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the object.

Does the escape velocity of a planet depend on its size?

Yes, the escape velocity of a planet is directly proportional to its mass and inversely proportional to its radius. This means that larger planets will have a higher escape velocity than smaller planets.

Can escape velocity be exceeded?

Yes, escape velocity is the minimum speed required to escape a planet's gravitational field. If an object has a higher speed, it will leave the planet's orbit and continue into space.

Is escape velocity the same for all objects on a planet?

No, the escape velocity for an object depends on its mass and the distance from the center of the planet. Objects with a greater mass will require a higher escape velocity than smaller objects.

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