Very basic gravitation question

[miringains]

---

Members don't see this ad.

Why exactly can you set mg equal to Gmm/r^2

And for centripetal motion you can also set it equal to mv^2/r

I believe I might know, but I would just like to confirm

and lastly, when is the equation Gmm/r used? For potential energies? But when

 

[brothersport]

F=ma.

The acceleration due to gravity on Earth is g. So in this case, a=g for any object on or near the surface of the Earth.

For uniform circular motion, you need to think in terms of calculus. The velocity magnitude is constant, but it changes direction.

acceleration = velocity * (dθ/dt)

We know that dθ = ω dt.

So a=vω=(v^2)/r

 

[illinichief89]

Why exactly can you set mg equal to Gmm/r^2

And for centripetal motion you can also set it equal to mv^2/r

I believe I might know, but I would just like to confirm

and lastly, when is the equation Gmm/r used? For potential energies? But when

GM/r^2 is equal to acceleration of a body at small distances. Factor out the smaller mass so: (m) (GM/r^2) = m(a). Since they both are describing the same gravitational force, we can set them equal to each other.

Also, for centripetal motion, if you look at v^2/r the units of v is (m/s). if you square that then you get (m^2/s^2). So if you divide that by r, which is measured in meters, then you get (m/s^2). That is the unit for acceleration. so m(v^2/r) =m(a)

Whenever you multiply a Force by a distance, you will get Energy, measured in joules. Since Gmm/r^2 is the gravitational force, then the gravitational potential energy is (Gmm/r^2) X r = (Gmm/r). Since we already said Gmm/r^2 can be approximated to mg, then mgh is the gravitational potential energy(where h is equivalent to r).
So: mgh = (Gmm/r)

 

[StilgarMD]

how does an object get further off the ground (more h, and more r), yet have one express of PE drop and the other raise