# Rocket trip.

• A
• MathematicalPhysicist

#### MathematicalPhysicist

Gold Member
I have this exercise which I don't understand its solution.

I am attaching both the exercise and its solution.

It's problem 2: rocket trip.
What I don't understand in the solution they write that:
"and from this you get:
$$\omega < \frac{1}{R}\sqrt{1-2m/R}$$
$$R>3m$$

Now, if I plug in this ##\omega## I get: ##m/R^2<\sqrt{1-2m/R}## and squaring both sides and multiplying by R^4 I get: ##m^2<R^4-2mR^3## ; if ##R>3m## then ##R^4-2mR^3>27m^4## and this is greater than ##m^2## whenever ##m^2>1/27##; from what does this follow I don't see, can't ##m^2## be less than ##1/27##?

#### Attachments

• sol6.pdf
118.1 KB · Views: 273
• ex6.pdf
162.9 KB · Views: 272
plug in ##\omega = m/R^3##

The solution appears to be wrong here. The correct equation you should get from the geodesic equation and the Christoffel symbols is ##\omega^2 = m / R^3## (note the squared ##\omega##), which you should recognize as Kepler's Third Law (note that this law turns out to hold exactly even in the relativistic case for this particular problem). This gives ##\omega = \sqrt{m / R^3}##; if you plug that into the other equation you should get ##R > 3m##, as expected.

MathematicalPhysicist
Btw, please note that it's highly preferred to post equations directly in LaTeX instead of attaching PDFs. I can understand not wanting to have to copy the entire problem and solution in, but just including a couple of the other equations would have been helpful (and in fact might have helped you to answer your own question).

Also, is this problem and solution available online? If so, links are better than PDFs.

berkeman