Satellite Slingshot manoeuvre question

[Crosshash]

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

Homework Statement

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.

Homework Equations

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

where

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

$l = V_{0}b$

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

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.

 

[Crosshash]

an unfortunate bump :(

 

[thomate1]

Hello! It looks like you have made a good attempt at solving this problem. However, your values for closest approach distance and maximum speed do seem a bit off. Here are a few things to consider:

1. Check your units: Make sure all of your units are consistent throughout your calculations. It looks like you have used meters for distance, but your units for mass and velocity are in kilograms and meters per second, respectively. This may be causing some discrepancies in your final values.

2. Consider the direction of the deflection: In your attempt at solving for the deflection angle, you have assumed that the angle is negative and that the sine of the angle is equal to -1. However, this may not be the case depending on the direction of the satellite's motion and the location of the planet. It's important to carefully consider the direction of the deflection when solving this problem.

3. Check your calculations: It's always a good idea to double check your calculations to make sure you haven't made any errors or typos. It's possible that a small mistake in your calculations could lead to significantly different values for closest approach distance and maximum speed.

Overall, it looks like you have a good understanding of the equations and concepts involved in this problem. I would recommend going back and carefully checking your units and calculations to see if you can identify any errors. Also, make sure to consider the direction of the deflection when solving for the angle. Keep up the good work!