big G = G = 6.674E-11 N(m/Kg)^2 Mass of earth = Me = 5.972E24 Kg Radius of earth = Rsurface = Rs = 6.371E6 metres THIS IS NOT HOMEWORK Okay so I was interested to know how you would work out the distance you'd need drop an object from (above the earths surface), for it to reach a given speed (neglecting air resistance). Here's what I've got so far: We know Force = F = G *(Mass of object 1 * Mass of object 2)/radius^2 = G * Mm/r^2 F = mass * acceleration = m*a F/m = a a = G * m/r^2 This is an equation for the acceleration of the object at any distance away from the centre of the earth (considering earth as a point mass for distances < Rs), taking into account the inverse square law of gravity. Right now it would be handy to have a in terms of t, so we could integrate to get an equation for v = velocity (units m/s). But instead we have it in terms of r (units metres). So I thought I might be able to integrate a in terms of r away to get something with the units (m/s)^2, which I thought might give us v^2. which we could square root to get an equation for v in terms of r? But I wasn't sure if that is correct/if I had overlooked some concept which means it isn't giving me what I wanted. Here's the integration I did: a = G * Me/r^2 v^2 = G * (integralOf(m/r^2)dr for the range lower limit = Rs to upper limit = r) = G * ([-(Me/r)] for the range Rs to r) so v^2 = G * (-Me/r - (Me/Rs) = G * Me*(r - Rs)/(r * Rs) so v = Sqroot( G * Me *(r - Rs)/(r * Rs) ) with a rearrangement ( (r*Rs) * v^2 )/(G*Me) - r = -Rs r * ( (Rs*v^2)/(G*Me) ) - 1 ) = -Rs r = -Rs/( ( (Rs*v^2)/(G*Me) ) - 1 ) ) Now I have tested this out and it wasn't correct (too far out to be correct when I compared it for the constant acceleration model given by the suvat equations). But I did notice that if you do this (using suvat equations): a = 9.81 acceleration s = r = displacement in metres u - start velocity = 0 m/s (metres per second) v^2 = 0^2 + 2*a*s v = Sqroot(2*a*s) you obviously get the right answer for constant acceleration, but what they're doing here is multiplying a by a distance vector to get v^2, which makes me think my decision to integrated above should have been okay (since integration find the area under a curve, which is like multiplication of the axes). Then I noticed that if you divide the right hand side of that suvate equation by Sqroot(2), you get extremely similar answers to my previous equation: v = Sqroot( G * Me *(r - Rs)/(r * Rs) ) So I multiplied the RHS of my variable acceleration equation by Sqroot(2) and it seems to be right. But I'm not sure where I went wrong in the first place! New equation: v = Sqroot(2 * G * Me *(r - Rs)/(r * Rs) ) So I do the rearrangement for r again and get (thinking it will work this time): r = -Rs/( ( (Rs*v^2)/(2*G*Me) ) - 1 ) ) Now it works! That value of r is including the earths radius, so you need to minus 6.371E6 to get the distance about the earth surface :). What I actually want help with READ :P: BUT I'm still not sure where that factor of 2 came from. I think it may have come from the integration, but the power of r in my initial equation for acceleration, was -2, which you increase to -1, then divide by the new power (being -1, not 2). Any clarification will be appreciated. Thanks! EDIT: Okay so I thought I'd also have a go at trying to find out the true gravitational potential energy, considering the inverse square nature of gravity. I just subbed the equation for v^2 into KE = 1/2 * m * v^2 :). So true GPE = 1/2 * m * (2 * G * Me *(r - Rs)/(r * Rs) ). If you want it relative to the centre of the earth make Rs 0, or if just for when the object hits the earths surface make Rs = 6.371E6 :).