# Shooting up a rocket

1. Nov 22, 2011

### Passionflower

Assume we are on a non rotating planet and shoot a rocket straight with a given velocity (smaller than the escape velocity) and we want to calculate using the Schwarzschild solution the coordinate and proper time, both for an observer on the planet and rocket, between lift off and return to the planet.

Let's take a very basic (but of course unrealistic) example:

The Schwarzschild radius rs = 1
The surface of the planet rp = 2
The velocity of the rocket v = 0.5

Here are my results:
For the apogee of the rocket I calculate: r=3
For the coordinate time I calculate: 14.90142209
For the proper time of a clock on the planet surface I calculate: 10.53689661
For the proper time of a clock on the rocket I calculate: 11.29502440

The question in this topic is about the approach used in:
"General Relativity and the Einstein Equations"
Choquet-Bruhat - Oxford, 2009

In chapter 9, page 87 we have:

http://img42.imageshack.us/img42/9963/formulae.png [Broken]

She uses the energy to calculate the proper time of a clock on the rocket. But I do not seem to get it to work.

If I calculate the Energy for the rocket for an observer at infinity I get: 0.8164965809

Since the rocket is free falling this energy must be a constant in the integral.

When solving this integral for the proper time on the rocket with this energy I get 9.125220141 which is less than what I calculate for the proper time on a clock on the surface of the planet.

Where do I go wrong or misunderstand?

Last edited by a moderator: May 5, 2017
2. Nov 23, 2011

### Passionflower

We can calculate the proper time for the rocket using:
$$\Large \int _{{\it r_i}}^{{\it r_o}}\!{\frac {1}{\sqrt {{\frac {{\it r_s}}{r}}-{ \frac {{\it r_s}}{{\it r_0}}}}}}{dr}$$
We can also express it in terms of energy and then we do not need the apogee at all:
$$\Large \int _{{\it r_i}}^{{\it r_o}}{\frac {1}{\sqrt {{E}^{2}-1+{\frac {{\it r_s}}{r}}}}}{dr}$$
(rs is the Schwarzschild radius and r0 is the apogee of the rocket and ri and ro are the r-coordinate ranges)

After multiplying by two both give a result of 11.29502440

3. Nov 24, 2011

### Passionflower

The formula in Choquet-Bruhat's book uses this principle:
$$\Large {\frac {dt}{d\tau}}=E \left( 1-{\frac {{\it r_s}}{r}} \right) ^{-1}$$
Which is correct as it is widely referenced in the literature.

But somehow it does not add up.

4. Nov 24, 2011

### Passionflower

And coordinate time can also be expressed in terms of energy:
$$\Large \int _{{\it ri}}^{{\it ro}}E \left( 1-{\frac {{\it rs}}{r}} \right) ^{-1}{\frac {1}{\sqrt {{E}^{2}-1+{\frac {{\it rs}}{r}}}}}{dr}$$

5. Dec 2, 2011