4 ms·

I was wondering about the necessary delta-v to do that vs. throwing it into the sun and found this [1] interesting table. Quite informative, it seems like the

I was wondering about the necessary delta-v to do that vs. throwing it into the sun and found this [1] interesting table. Quite informative, it seems like the naive approach of going straight into the sun is much more expensive (dv of 24.0 km/s) than escaping the solar system (dv of 8.8 km/s). But, it seems that there is a trick where you use 8.8km/s to almost escape the solar system, then turn around with very little dv cost and plunge into the sun. [1] https://en.wikipedia.org/wiki/Delta-v_budget#Interplanetary https://en.wikipedia.org/wiki/Delta-v_budget#Interplanetary

mnw21cam 7y ago
Yes, that's basic orbital mechanics. It takes longer that way though. There are many examples in orbital mechanics where you can trade efficiency for patience.
mannykannot 7y ago
The general rule for circular orbits is that the escape velocity is √2 times the orbital velocity. At no point on the esacape trajectory can the object's speed fall below √2 times that of a circular orbit at that distance (or else the object would not escape.) At whatever distance you decide to set the controls for the heart of the sun, you must kill its angular velocity with respect to the sun (because, if it has more than a slight angular momentum, it will follow an elliptical orbit that goes around the sun.) Therefore, at every point on the minimal escape trajectory, the delta-v to redirect the payload into the sun is the escape velocity at that distance. With your strategy, the cost of sending the payload into the sun asymptotically decreases towards the cost of sending it on an escape trajectory.
pontifier 7y ago
I was amazed the first time I heard about the Interplanetary Transport Network. https://en.m.wikipedia.org/wiki/Interplanetary_Transport_Network https://en.m.wikipedia.org/wiki/Interplanetary_Transport_Net...

mnw21cam 7y ago

Yes, that's basic orbital mechanics. It takes longer that way though. There are many examples in orbital mechanics where you can trade efficiency for patience.

mannykannot 7y ago

The general rule for circular orbits is that the escape velocity is √2 times the orbital velocity. At no point on the esacape trajectory can the object's speed fall below √2 times that of a circular orbit at that distance (or else the object would not escape.) At whatever distance you decide to set the controls for the heart of the sun, you must kill its angular velocity with respect to the sun (because, if it has more than a slight angular momentum, it will follow an elliptical orbit that goes around the sun.) Therefore, at every point on the minimal escape trajectory, the delta-v to redirect the payload into the sun is the escape velocity at that distance. With your strategy, the cost of sending the payload into the sun asymptotically decreases towards the cost of sending it on an escape trajectory.

pontifier 7y ago

I was amazed the first time I heard about the Interplanetary Transport Network. https://en.m.wikipedia.org/wiki/Interplanetary_Transport_Network https://en.m.wikipedia.org/wiki/Interplanetary_Transport_Net...