# Distance from earth an object would have to fall from to reach speed x

1. May 11, 2013

### cawthorne

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 :).

Last edited: May 11, 2013
2. May 11, 2013

### arildno

"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."

Sure. But remember that for EVERY dimensionless constant "k", the quantity k*v^2 has the right dimensions!

Thus, without knowing which "k" is the right one, you won't get the right answer.

The factor 1/2 is the one appearing in the quantity called "kinetic energy",

3. May 11, 2013

### cawthorne

Thanks for the reply. I'm still not sure where I went wrong. I'd have thought that is my acceleration equation was correct, then as long as I integrated everything properly my v^2 equation should still be correct, because I'd have though that the division part of integrating would have sorted out any constant factors?

Also where is the 1/2 that is appearing you speed of? It is a factor of 2 that appears, which is the reciprocal of 1/2. But I haven't used the KE equation. I know that KE = 1/2 m v^2 can be derived using F = ma, which I did use.

Any more explanation would be appreciated :).

4. May 11, 2013

### 256bits

Step by step

a = GM/r$^{2}$

a dr = GM dr / r$^{2}$

a = dv/dt

Subst and re-arrange

vdv = GM/r$^{2}$ dr

Integrate, yielding

v$^{2}$/2 = -GM/r

v$^{2}$ = -2GM/r

Now you have the 2 ( and the square root of 2 )

Take your limits or r from r1 to r2 where r1>r2 and you should come up to your good equation.

5. May 12, 2013

### cawthorne

Thank you very much! So my mistake was to presume that I could jump from:

a dr ----> v^2

because the units matched up :P. Well this was fun and definitely helped me brush up on my maths :)!!!!!

Regards, Greg.