Gravitational potential energy

[alex36]

Suppose the mass of planet is" M" and there is body in its surface whose mass is "m" and the field strength is "g" . If the body is thrown 1800 m then Gravitational Potential energy = mg(1800). My question is why cant we use formula GPE= GMm/x ? This is also the formula for gpe but why cant we apply it in this condition?

 

[robphy]

What is the physical interpretation of x?

 

[alex36]

What is the physical interpretation of x?

distance from the surface of planet

 

[Jonathan Scott]

The assumption that the field strength is constant is a simplification which applies when the height involved is small compared with the distance to the centre of the source object.

In this case, we can simply use the approximate formula mgh for the energy when mass m is moved in field g = GM/x^2 through height h. The -GMm/x formula is correct only if x is the distance to the centre of the planet. In that case the change in potential energy can also be written accurately as (-GMm/(x+h)) - (-GMm/x) which is approximately the same as mgh provided that h is small compared with x.

 

[alex36]

The assumption that the field strength is constant is a simplification which applies when the height involved is small compared with the distance to the centre of the source object.

In this case, we can simply use the approximate formula mgh for the energy when mass m is moved in field g = GM/x^2 through height h. The -GMm/x formula is correct only if x is the distance to the centre of the planet. In that case the change in potential energy can also be written accurately as (-GMm/(x+h)) - (-GMm/x) which is approximately the same as mgh provided that h is small compared with x.

Isn't it (-GMm/x-(-GMm/(x+h))? because energy we get is negative from your form of euation . does it matters?

 

[Jonathan Scott]

I'm assuming that something is being thrown upwards, so a positive amount of potential energy being given to the small mass.

The Newtonian potential energy in the form -GMm/x is relative to infinite separation, so it gets lower the closer one gets to the source. To compare it at two different heights you subtract the potential energy values. In this case, the higher energy is the the one which involves (x+h), which is less negative, making the difference positive in the form I gave originally.

Of course, the mgh form also needs care with the sign. This is the potential energy lost when the small mass falls a distance h in the same direction as the field g, so it is also the same as the potential energy which has to be given to the small mass to move it a distance h in the opposite direction to the field g.

 

[hackhard]

. If the body is thrown 1800 m then Gravitational Potential energy = mg(1800). My question is why cant we use formula GPE= GMm/x ?

a simple concept-
there is no such thing as "absolute potential energy"
when you say a body has 20m/s speed you also need to say -- in which frame ?
when you say a bode has 20J potential energy you also need to define--- which point have you assumed as ground potential (0J)
mgh gives gravitational potential energy (due to earths gravity) assuming earth surface to be at ground potential (at 0J)
GPE= GMm/x gives gravitational potential energy (due to earths gravity) assuming infinity to be at ground potential (at 0J)
both are correct provided you also mention which point have assumed to be at ground potential
simply saying " potential energy of a body is 200J" is non sense
you say " potential energy of a body is 200J wrt point A"
so a body can have all real values as potential energy at same point of time but a unique one wrt to a choice of ground

 

[alex36]

I just checked . Answer will have same value but different sign . Am I correct?

 

[Jonathan Scott]

I just checked . Answer will have same value but different sign . Am I correct?

The answer should have the same sign whichever way you calculate it.

One way, the energy is GMmh/x(x+h), which is approximately m (GM/x^2) h, where GM/x^2 is the magnitude of g, and the other way is mgh where g is the magnitude of the field and h is assumed to be upwards.

If you want to be accurate about signs, the energy given to the mass in the mgh form is actually -mg.h if the field and height are described by vectors, because the force being applied to the mass is in the opposite direction to gravity, but the displacement through which it acts is in the forward direction, so we have -m(-GM/x^2) h = m (Gm/x^2) h as before.

 

[alex36]

The answer should have the same sign whichever way you calculate it.

One way, the energy is GMmh/x(x+h), which is approximately m (GM/x^2) h, where GM/x^2 is the magnitude of g, and the other way is mgh where g is the magnitude of the field and h is assumed to be upwards.

If you want to be accurate about signs, the energy given to the mass in the mgh form is actually -mg.h if the field and height are described by vectors, because the force being applied to the mass is in the opposite direction to gravity, but the displacement through which it acts is in the forward direction, so we have -m(-GM/x^2) h = m (Gm/x^2) h as before.

Thank you so much :)