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

[JFonseka]

Homework Statement

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.

Homework Equations

v = \sqrt{2GM/R}

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

 

[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

 

[BlackWyvern]

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

Ep = -GMm/r

\DeltaEp = Ep_f - Ep_i
Ep_f = -GMm/3r
Ep_i = -GMm/r

-GMm/3r - -GMm/r = \DeltaEp
-GMm/3r + GMm/r = \DeltaEp
-GMm/3r + 3GMm/3r = \DeltaEp
2GMm/3r = \DeltaEp

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.

 

[JFonseka]

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

Ep = -GMm/r

\DeltaEp = Ep_f - Ep_i
Ep_f = -GMm/3r
Ep_i = -GMm/r

-GMm/3r - -GMm/r = \DeltaEp
-GMm/3r + GMm/r = \DeltaEp
-GMm/3r + 3GMm/3r = \DeltaEp
2GMm/3r = \DeltaEp

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

If we solve that, we see that the simple solution seems to be correct.
So someone is wrong. <_<
I see no flaw in my logic, I don't see their logic. So I feel that the answers you are given are incorrect.

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.

 

[mgb_phys]

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

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

 

[JFonseka]

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

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