# Minimum escape velocity of a projectile required to rise a certain height

1. Aug 26, 2007

### JFonseka

1. The problem statement, all variables and given/known data
A projectile is launched from the surface of a planet (mass = M, radius = R). What minimum launch speed is required if the projectile is to rise to a height of 2R above the surface of the planet? Disregard any dissipative effects of the atmosphere.

2. Relevant equations

v = $$\sqrt{2GM/R}$$

3. The attempt at a solution

So we know that for a projectile to rise to a height of R above the planet, the equation above will suffice, however the projectile in this question has to rise to a height of 2R, that is R + 2R, so 3R. I thought you simply replace R by 3R, to get v = $$\sqrt{2GM/3R}$$

But that's not the case, the multiple choices does not have that as one of the answers. The answers listed were...

a) $$\sqrt{\frac{4GM}{3R}}$$
b) $$\sqrt{\frac{8GM}{5R}}$$
c) $$\sqrt{\frac{3GM}{2R}}$$
d) $$\sqrt{\frac{5GM}{3R}}$$
e) $$\sqrt{\frac{GM}{3R}}$$

I think the most sensible looks to be GM/3R but I'm probably wrong, how should I go about doing this problem?

Thanks

2. Aug 27, 2007

### mgb_phys

Potential energy is = GMm/R
Kinetic Energy = 1/2 m v^2
Where R is distance form centre of planet with mass M and m is the mass of the object.
You are going from R=r to R=3r so change in potential energy is
PE = GMm/r - GMm/3r = 2GMm/3r

Set this equal to KE, cancel m and solve for v

3. Aug 28, 2007

### BlackWyvern

e would be the energy to maintain an orbit at 3R.

Ep = -GMm/r

$$\Delta$$Ep = Epf - Epi
Epf = -GMm/3r
Epi = -GMm/r

-GMm/3r - -GMm/r = $$\Delta$$Ep
-GMm/3r + GMm/r = $$\Delta$$Ep
-GMm/3r + 3GMm/3r = $$\Delta$$Ep
2GMm/3r = $$\Delta$$Ep

2GMm/3r = mv2
- Is the condition for escaping Earth's gravity.

Edit:
2GMm/3r = (1/2)mv2
4GM/3r = v2
$$\sqrt{\frac{4GM}{3r}}$$

I forgot the 1/2, lol.

Just for the record, second poster made two big mistakes. You don't subtract initial from final to get delta. AND, the formula contains a negative sign.

Last edited: Aug 29, 2007
4. Aug 28, 2007

### JFonseka

Nope, they are quiz answers, and it would have been pointed out by now by some other student or the lecturer if the answers were wrong, the correct answers was the 4GM/3R

When in fact we use the 2nd poster's method it turns out to be right.

5. Aug 28, 2007

### mgb_phys

I did check the answer but it would have taken all the fun out of it to tell you!

6. Aug 28, 2007

### JFonseka

Well it wasn't that hard to figure it out after that, divide by half, divide by m and take the square root, lol