Question

Explain formula F=mv*v/r

Answer 1

This follows directly from the experiment Sketching a satellite orbit and predicting its period. The mathematics follows directly from the sketch produced in that experiment, reproduced below. This is Newton’s method. 

It relies crucially on the crossed chords theorem for a circle, which should be given. 

The circle represents the orbit of a satellite of radius R, moving with speed v. The satellite moves from A to B in a time t. Without a force the satellite would have moved to K at constant speed. 
 
Now 'switch-on' gravity and the satellite will fall a distance h in the same time from the tangent from A to the point B. It doesn’t matter whether you let it fall from A first, and then continue in the tangential direction, or vice versa. Anyone who objects that the fall from K to B is not along the radius should look at their scale diagram again: it is almost impossible to see the difference between h and the radial drop. [You may need to talk about 'in the limit'.] 
 
From the crossed chords property, h(2R-h) = x2 
but 2R>> h therefore 2h = x2 and so h = x2/2R (equation 1) 
now x = AK which is almost the arcAB = vt (equation 2) 
 
Combining 1 and 2, h = (vt)2/2R (equation 3) 
h is the vertical fall and so using s = 1/2 at2 = h (equation 4) 
 
Then from (3) and (4) 1/2at2 = (vt)2/2R leading to a = v2/R 
Using F = ma then F = mv2/R 
 
The same holds for the motion at all places round the circle. The vertical is always taken to mean the direction from the satellite to the centre of the attracting body. 
 

Answer 2

This follows directly from the experiment Sketching a satellite orbit and predicting its period. The mathematics follows directly from the sketch produced in that experiment, reproduced below. This is Newton’s method. 

It relies crucially on the crossed chords theorem for a circle, which should be given. 
The circle represents the orbit of a satellite of radius R, moving with speed v. The satellite moves from A to B in a time t. Without a force the satellite would have moved to K at constant speed. 
 
Now 'switch-on' gravity and the satellite will fall a distance h in the same time from the tangent from A to the point B. It doesn’t matter whether you let it fall from A first, and then continue in the tangential direction, or vice versa. Anyone who objects that the fall from K to B is not along the radius should look at their scale diagram again: it is almost impossible to see the difference between h and the radial drop. [You may need to talk about 'in the limit'.] 
 
From the crossed chords property, h(2R-h) = x2 
but 2R>> h therefore 2h = x2 and so h = x2/2R (equation 1) 
now x = AK which is almost the arcAB = vt (equation