# Gravitation energy 3

1. Jun 22, 2014

### Karol

1. The problem statement, all variables and given/known data
A mass m is sent from earth into an orbit. the period is T.
What is the energy required to put it into orbit. express it with T, m, M, R(radius of earth)

2. Relevant equations
$$E=\frac{1}{2}mv^2-\frac{GMm}{r}$$
$$2\pi r=vT$$

3. The attempt at a solution
I expressed the radius of the orbit r and the velocity:
$$r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}$$
$$v=\sqrt[3]{\frac{2\pi GM}{T}}$$
Those results are correct according to the book. the energy required:
$$E=\frac{1}{2}mv^2-\frac{GMm}{r}-\left(-\frac{GMm}{R}\right)$$
$$E=m\frac{\sqrt[3]{\frac{4\pi^{2}G^{2}M^2}{T^2}}}{2}-GMm\left(\sqrt[3]{\frac{4\pi^2}{GMT^2}}-\frac{1}{R}\right)$$
$$E=m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{2T^2}}-m\sqrt[3]{\frac{G^{3}M^{3} 4\pi^2}{GMT^2}}+\frac{GMm}{R}$$
$$E=m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{2T^2}}-m\sqrt[3]{\frac{G^{2}M^{2} 4\pi^2}{T^2}}+\frac{GMm}{R}$$
The answer in the book is:
$$E=GMm\left( \frac{1}{R}-\sqrt[3]{\frac{\pi^2}{2GMT^2}}\right)$$
I can reach to that form, but this answer includes only the first:
$$\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{2T^2}}$$
And the last elements in my result.

2. Jun 22, 2014

### Karol

ff

I solved it using the energy in an orbit:
$$E=-\frac{GMm}{2r}$$

3. Jun 23, 2014

### dauto

The first and second terms of your solution can be added together and are identical to the second term of the books solution.

4. Jun 25, 2014

### Karol

My solution:
It's not true that the first and second terms of my solution make the term in the book:
$$m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{2T^2}}-m\sqrt[3]{\frac{G^{2}M^{2} 4\pi^2}{T^2}}=\frac{1}{8}m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{T^2}}-64m\sqrt[3]{\frac{G^{2}M^{2} \pi^2}{T^2}}=-63\frac{7}{8}m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{T^2}}$$
And the term in the book gives:
$$GMm\sqrt[3]{\frac{\pi^2}{2GMT^2}}=m\sqrt[3]{\frac{G^{3}M^{3}\pi^2}{2GMT^2}}=m\sqrt[3]{\frac{G^{2}M^{2}\pi^2}{2T^2}}$$
And it's only the first term in my answer