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Spacecraft acceleration using a planet's gravity

When a spacecraft travels to the outer solar system it picks up speed by using the gravity of planets along the way. In fact they sometimes would whizz by the earth a few times to pick up speed before continuing their journey. It does work, but I don't understand how it can work.

Suppose a spacecraft is heading to saturn. It approaches jupiter and gradually picks up speed as it nears the planet. However, as it pulls away from jupiter it must now fight the gravity of jupiter which would slow it down. Give the "conservation of energy" and that the distances approaching and leaving the planet are the same, how can the space craft gain energy?

In addition, these spacecraft normally change direction which would use enegy.
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Eirman
Asked:
Eirman
2 Solutions
 
sdstuberCommented:
see if this helps explain things

http://en.wikipedia.org/wiki/Gravitational_slingshot
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NovaDenizenCommented:
Let's say that you have a stationary planet (forget about the sun, moons, the rest of the planets, etc) and a small ship on a trajectory that will pass near the planet.  Assuming the ship is not going so slowly that it is already in orbit around the planet, the ship will follow a hyperbolic path around it, changing its direction but in the end not affecting its velocity.  Changing the direction of travel of the ship will impart an equal momentum change of the planet, which means the planet will no longer be perfectly stationary.

The speed of the ship will be a simple function of its distance from the planet.  When the ship is approaching the planet from one million km out, it will have the same speed as when it is departing the planet and one million km away.  This is obvious from conservation of energy.

Now remember that the concept of being stationary is not absolute.  From a moving planet's frame of reference, the ship will approach at a certain velocity/distance function, and depart with the same velocity/distance function.  

Let's say that the ship is moving orthogonally towards the path of the planet at 10 km/s at a distance of 1 million km, and the planet is moving at 50 km/s, the planet sees the ship coming at a relative speed of sqrt(50^2 + 10^2) = 51km/s.  The planet sees the ship whip around itself and start departing on a typical hyperbolic trajectory.  When the ship reaches a distance of 1 million km from the planet, it will be moving out at a relative speed of  51 km/s.  But to us, we see the ship moving now at sqrt(100^2 + 10^2) = 100.5 km/s.
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EirmanAuthor Commented:
Thanks Nova - I'll dig out the old physics books so I can fully understand your explanation.
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aburrCommented:
The short and oversimplified explanation follows.
The sling short method adds angular momentum to the spacecraft. The planet looses a corresponding amount of angular momentum. (But because it has so much the reduction of rotational speed is unnoticeable.) Because of a relationship between angular momentum and kinetic energy, the net result is a transfer of kinetic energy from the planet to the spacecraft.
Warning: If too many spacecraft use Jupiter for a slingshot, its day will be affected. (Exercise for the reader: If 25 slingshort events pet year involve Jupiter, how many centuries will have to elapse before the day of Jupiter decreases by 1 second?}
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NovaDenizenCommented:
Will it really affect the day length of the planet, or will it affect the planet's year length?  Unless the spaceship is large enough to cause detectable transient tidal forces on the planet, I don't think the rotation of the planet will be affected.
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ozoCommented:
It doesn't affect the day length of the planet, it affects the year length of the planet.

The tidal forces won't be detectable on the spacecraft, much less the planet.
And the effect on year length of the planet is not even detectable,
Anyway, the undetectable effect of tides is more likely to slow down the spacecraft and speed up the planet
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