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