## chris215 one year ago Va is the orbital speed of a satellite at a distance x from the centre of the Earth. If the satellite is at a distance 2x from the centre of the Earth and with the speed of Vb, which of the following statements is correct?

1. chris215

va = vb^2?

2. Astrophysics

|dw:1435735326495:dw| we need to use the following formula $\huge v_{escape}^2 = \frac{ 2GM }{ r }$ so we have $\large v_a^2 = \frac{ 2GM }{ x }$ and $\large v_b^2 = \frac{ 2GM }{ 2x } = \frac{ GM }{ x }$

3. Astrophysics

So now we have $\large v_a = \sqrt{\frac{ 2GM }{ x }} \implies \sqrt{2}\sqrt{\frac{ GM }{ x }}$ and we have $\large v_b = \sqrt{\frac{ GM }{ x }}$ therefore we can say $\huge v_a = \sqrt{2}v_b$

4. Michele_Laino

for mechanical stability, we can write: $\Large \begin{gathered} m\frac{{v_a^2}}{x} = \frac{{GMm}}{{{x^2}}} \hfill \\ \hfill \\ m\frac{{v_b^2}}{{2x}} = \frac{{GMm}}{{{{\left( {2x} \right)}^2}}} \hfill \\ \end{gathered}$ where m is the mass of our satellite, and M is the mass of Earth. Now simplifying both equations, I get: $\Large {v_a} = \sqrt {\frac{{GM}}{x}} ,\quad {v_b} = \frac{1}{{\sqrt 2 }}\sqrt {\frac{{GM}}{x}}$ So, what can you conclude?

5. Astrophysics

Why did you do that @Michele_Laino ?

6. Michele_Laino

It is another way to solve that question @Astrophysics