Minimum velocity to throw a object to another planet

[Yalanhar]

Nevertheless the answer you are aiming for is wrong.

If D is small, then as mentioned above you have a very different problem.

I'm not sure what else to suggest.

I'll try what jbriggs444 suggested about adding energies

 

[TSny]

The answer I got is:

$v_0^2=2GM\frac{(D+R)^2}{R(D+2R)(D+3R)}$.

Liable to my own algebraic error of course. I hope that posting this is allowed, since OP showed that he has the intended final answer available.

Yes, that's the answer I get also, assuming that D is the distance between surfaces.

 

[Yalanhar]

The answer I got is:

$v_0^2=2GM\frac{(D+R)^2}{R(D+2R)(D+3R)}$.

Liable to my own algebraic error of course. I hope that posting this is allowed, since OP showed that he has the intended final answer available.

May I ask how?

 

[Highway_Dylan]

May I ask how?

Try what I (and also others) wrote in post #28. You can get the answer by using conservation of energy.

 

[PeroK]

The answer I got is:

$v_0^2=2GM \frac{(D+R)^2}{R(D+2R)(D+3R)}$

Latex sorted. We're all in agreement then.

@Yalanhar

In any case, if we are taking D to be the distance between the surfaces (which is really odd in a gravitational problem), then I would let $d = D +3R$ be the distance between the centers. Then work with $d$, as it is simpler and more standard, only converting to D for the final answer.