# Satellite Slingshot manoeuvre question

1. Nov 20, 2008

### Crosshash

I've already attempted to work everything out by the way but my answers seem a little off.

1. The problem statement, all variables and given/known data
A satellite performs a sling shot manoeuvre around a planet. The mass of the planet is $6.00 * 10^{24} kg$ and the mass of the satellite is $7.00 * 10^{2} kg$. The satellite approaches the planet from a great distance with relative speed $2 * 10^{3} ms^{-1}$ and impact parameter $6.0 * 10^{7} m$. Determine the closest approach distance, the maximum speed of the satellite relative to the planet and the angle through which it is deflected.

2. Relevant equations
$1/r = \alpha \sqrt {2c/l^{2} + \alpha^{2}} \hspace {5 mm}sin (\theta + \theta_{o})$

where

$\alpha = {G(M_{1}+M_{2})}/l^{2}$

$l = V_{0}b$

$c = 1/2 V_{0}^{2}$

3. The attempt at a solution

I've attempted to solve the whole question, but I think my values for the closest approach distance and maximum speed are quite wrong.

$M_{planet} = 6.00 * 10^{24} kg$
$M_{satellite} = 7.00 * 10^{2} kg$
$V_{0} = 2*10^{3} ms^{-1}$
$Impact Parameter = b = 6.00 * 10^{7}m$

Deflection Angle
$\sqrt {1 + 2c/l^{2}\alpha^{2}} \hspace {5 mm} sin \theta = -1$

$\alpha = {G(M_{1}+M_{2})}/l^{2}$

$l = V_{0}b$

$c = 1/2 V_{0}^{2}$

$c = (2 * 10^{3})^{2}/2 = 2*10^{6}$

$l = (2 * 10^{3}) * (6.0*10^{7}) = 1.2*10^{11}$

$\alpha = G(6.00*10^{24} + 7*10^2)/l^{2} = G(6*10^{24})/(1.44*10^{22}) = 4.002*10^{14}/1.44*10^{22} = 2.78*10^{-8}$

$\sqrt {1 + 2c/l^{2}\alpha^{2}} = \sqrt {1 + \dfrac{2*(2*10^{6})} {(1.2*10^{11})^{2} (2.78*10^{-8})^{2}}$

$\sqrt {1 + 0.36} = \sqrt {1.36}$

$sin \theta = {\dfrac{-1} {\sqrt 1.36}}$

$\theta = -59.04$

Angle of Deflection = 118.08

Unless anyone can point out a problem here, this seems fine.

Closest Approach Distance

$\dfrac {1}{r_{min}} = \alpha \sqrt {\dfrac {2c}{l^{2}} + \alpha^{2}}$

$\dfrac {1}{r_{min}} = 2.78*10^{-8} * \sqrt {\dfrac{4*10^{6}}{(1.2*10^{11})^{2}} + (2.78*10^{-8})^{2}}$

$\dfrac {1}{r_{min}} = 2.78*10^{-8} \sqrt {1.051*10^{-15}}$

$\dfrac {1}{r_{min}} = 9.01*10^{-16}$

$r_{min} = 1.11*10^{15}m$

This seems quite large, and it means that my speed at closest approach is very small.

Speed at closest approach

$\dfrac{v_{0}}{b} = \dfrac {1.2*10^{11}}{1.11*10^{15}} = 1.08*10^{-4}ms^{-1}$

This seems very small which is why i'm wondering if someone could tell me where I went wrong.

This is also the first time i've tried extensively using latex so pardon me if it's hard to read.

Thanks guys.

2. Nov 20, 2008

### Crosshash

an unfortunate bump :(