What is the main difference between the formulas for kinetic energy (KE) and gravitational potential energy (PE)?

Question

anonymous · Answer

kinetic energy is
$$ke=1/2 mv ^{2}    and   pe=mvh$$
for kinetic energy height is no variable but for PE it is. . .

ujjwal · Answer

@theyatin , a bit of error in your reply.
$$KE=\frac{1}{2}mv^2$$and $$PE=mgh$$

anonymous · Answer

oh oops i typed it in hurry srry. . .

anonymous · Answer

Ahm.. could you give us some context for this question? Asking for difference, when there is no apparent connection is really weird. 

Also, I'd argue the formula for grav. pot. Energy to be: $$E = -G {{m_1 m_2} \over {r}}$$

anonymous · Answer

well this is gravitational law for heavanly bodies. . . well i say dont consider it. . .

anonymous · Answer

You say: "Don't consider the one, that always holds, but take the approximation for special circumstances?" .. Well, you may be righ t and that's indeed what is required, but the main problem here is the lack of information/context and thus the question making no sense to begin with..

Could you help us out more, gberg50 ?

anonymous · Answer

common sense says so if he has mentioned kinetic energy that heavenly bodies dont have kinetic energy

anonymous · Answer

I would agree with MuH4hA

anonymous · Answer

@theyatin - why would you think "heavenly bodies" don't have kinetic energy? If such a body - for example an asteroid - would collide with, let's say the moon, I will claim that the size of the bang it makes is proportional to some number 1/2 m*v², where m is the mass of the asteroid and v is it's velocity relative to the moon. I'd further call that number "kinetic energy of the asteroid". 
Now why would you claim that this does not make sense with celestial bodies, but accept it with.. say a baseball?

anonymous · Answer

okay bro you took it way too much seriously heavenly body do not just have kinetic energy they have angular(circular) motion mostly so you can not take 1/2mv^2.
rest i dont want to argue. . .