Question

gravitational potential in a field due to point mass ?

Answer 1

\[v =GM/r\] What is the use of this equation

Answer 2

v=GM/r
what is this @nirmalnema

Answer 3

This is gravitational potential in a field due to point mass
WHat does that mean actually

Is this same as u=mgh ?

Answer 4

@triciaal

Answer 5

I don't remember this stuff but I looked it up just now for you
The gravitational potential (V) is the potential energy (U) per unit mass:
U = m V,
where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity. If the body has a mass of 1 unit, then the potential energy to be assigned to that body is equal to the gravitational potential. So the potential can be interpreted as the negative of the work done by the gravitational field moving a unit mass in from infinity.
In some situations, the equations can be simplified by assuming a field that is nearly independent of position. For instance, in daily life, in the region close to the surface of the Earth, the gravitational acceleration can be considered constant. In that case, the difference in potential energy from one height to another is to a good approximation linearly related to the difference in height:
\Delta U = mg \Delta h.

Answer 6

The potential V at a distance x from a point mass of mass M can be defined as the work done by the gravitational field bringing a unit mass in from infinity to that point:
V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x},

Answer 7

thanks @triciaal

Answer 8

where to use PE = mgh
and PE = GMm/r
@sidsiddhartha