Escape Velocity and planet’s gravitational field

[RoryP]

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

E_K= 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!

 

[LowlyPion]

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

E_K= 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 = Gm_p/r_p

Where r_p and m_p are the radius and mass of the dwarf planet Pluto

 

[RoryP]

ahhh sweet cheers! =]

 

[Dweirdo]

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