00:01
Okay, in this question we have a satellite of mass m, which is orbiting around the earth.
00:05
The earth has a mass, well, i'm just going to label the mass m -e.
00:09
Satellite has a mass m, and then the radius of the orbit is r.
00:17
And really, it's going to be the distance to the center of the earth.
00:20
That's going to be r.
00:22
So we are asked to express the kinetic energy of the satellite in terms of g, m, and r, where g is the universal gravitational constant.
00:31
So in order to do this, we are going to equate.
00:33
The centripetal force, so f is equal to mv squared over r, with the force of gravitational attraction between the satellite and the earth, which is g, m, m over r squared.
00:48
So we can say that mv squared over r is equal to g m m over r squared.
00:57
You can cancel out the r on the left hand side, but the squared on the right hand side.
01:02
And then we have mv squared, is equal to g m e, is equal to g m e m over r now we know that kinetic energy is half mv squared therefore if we just say half mv squared is equal to one over two g m over r we can just neaten that up by saying g m m over two r again where g is gravitation constant m is the mass of the earth m is the mass i really i should probably write that in lowercase.
01:37
M is the mass of the satellite and then r is the radius of the orbit.
01:43
And then for b, in terms of the given quantities, calculate the period of the orbit, or this, if you ask for a period for any orbit, it's going to be kepler's third law.
01:55
And we're going to, again, go back to equating these two equations.
01:58
So we have mv squared over r is equal to g m m over r squared and again we can cancel out squared with the r so we have mv squared is equal to g m e m over r okay so how do we then get time period into this well if we think about it this satellite is moving in a circle okay so its velocity is going to be the distance over the time taken.
02:34
So v, i'm doing the same green, v is equal to 2 pi r over t.
02:42
So we've now found a way to get time period into our equation.
02:47
So, that what we can also do is we can cancel the m's as well.
02:54
So we have v squared is equal to 2 pi r, oh, no, sorry, v squared is equal to gm, gm, over r and therefore because v is 2 pi r over t we can say v squared is 4 pi squared r squared over t squared because two squared is four pi squared is pi squared is r squared and t squared is t squared so that is v squared there v squared is equal to g m e over r all right and then we can just isolate t so we can then say isolating t we say t squared is equal to 4 pi squared r cubed over g m .e, therefore t is equal to the square root of all that.
03:51
T is equal to the square root of 4 pi squared r cubed over g m .e.
03:58
Okay.
03:59
And we are then, well, how does it depend on the mass of the satellite? and given that the mass cancels out, it doesn't depend at all on the mass.
04:09
Of the satellite.
04:12
Then c, show that gravity does zero work on the satellite whilst it is in circular orbit.
04:17
So circular orbit is key there.
04:19
If it's in circular orbit, then the radius does not change.
04:23
So, so that was, rather than two, that should say b, c, and that's changed the blue for a bit of clarity.
04:33
Gravity does no work.
04:35
Well, if force is equal to gmm, i should, just say g mem, mass of the earth, multiple mass of the satellites, over r squared.
04:52
And then work done is, well, that's going to be force times distance moves...