# Velocity of an satellite in an elliptical orbit

Consider some some satellite of mass $m$ around earth, traveling in an elliptical orbit about a focus (earth). Let's suppose at one point $A$ the satellite is a distance $a$ units away the focus at minimum distance, and at another point $B$ it is at a distance of $b$ from the focus, a maximum distance. Let points $A$ and $B$ be collinear with the center of the Earth. When the satellite is at a distance of $a$, it's speed is equal to $v_0$.

Calculate the speed of at the satellite a distance $b$ from the focus.

So first I approached this by the conversion of mechanical energy:

$$\frac{1}{2}mv_0^2-\frac{Gm_e m}{a}=\frac{1}{2}m v_b^2-\frac{G m_e m}{b}$$

I got,

$$v_b=\sqrt{v_0^2+2Gm_e(\frac{1}{b}-\frac{1}{a})}$$

But then I realized conservation of angular momentum gives,

$$mv_bb=mv_0 a$$

$$v_b=\frac{a}{b}v_0$$

I don't see how the two can be equivalent. Is one answer wrong? May someone please explain.

The semi-major axis equals to $\dfrac{a+b}{2}$ where $a<b$.

$$v^2=2GM\left( \frac{1}{r}-\frac{1}{a+b} \right)$$

At perigee ($r=a$),

$$v_a=\sqrt{\frac{2GMb}{a(a+b)}}$$

At apogee ($r=b$),

$$v_b=\sqrt{\frac{2GMa}{b(a+b)}}$$

Both total energy and angular momentum are conserved.

See the link here for further information.

The second answer is wrong because it is missing the angles from the vector products: https://en.wikipedia.org/wiki/Angular_momentum#Vector_.E2.80.94_angular_momentum_in_three_dimensions

• Sorry I don't think I expressed the problem well enough, in the problem it's obvious the angle between the velocity vector and the position vector is $\90$ degrees. – Ahmed S. Attaalla Apr 8 '17 at 5:43
• No, on an elliptical orbit this is not the case, see the difference between T and F on the following image: star-www.st-and.ac.uk/~fv/webnotes/ELLIPVEL.GIF – Photon Apr 8 '17 at 5:45
• Ok, looking at your picture, I think I understand your last statement. For the particular case of the two points A and B in the picture the second answer is correct as well. For this case both formulas should give the same numbers though they look very different. This is because quantities that appear in the equations are not independent. If you fix $v_0$, $a$ and $Gm_e$, you should be able to predict $v_b$ and $b$. Once you express both results in terms of the smallest number of independent quantities, the formulas should become the same. – Photon Apr 8 '17 at 5:57
• @photon At positions A and B the velocity of the satellite is at right angles to the straight line joining the centre of the Earth to the satellite. As the force is central, angular momentum is conserved. – Farcher Apr 8 '17 at 8:13
• @Farcher: That's what I wrote in my last comment. ;) – Photon Apr 8 '17 at 8:39