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Yes it matters- the Sun is the main provider of gravity for this consideration.

by dhimes 13y ago

Groxx 13y ago
orbit. gravity is canceled out by orbital velocity, as far as distance-from-sun is concerned. it costs Mars nothing to stay in orbit, and to go down to a -100m/s orbit or up to a +100m/s orbit costs the same amount of energy: (mass of mars) * (100m/s velocity change) if Mars and the Earth were not orbiting, I would completely agree. drop something from Mars and it'll land on Earth, and the reverse is not true. but they're not - drop something on Mars and it's just in Mars' orbit.
- mikeash 13y ago
  Delta-v requirements for transfers in space are generally symmetrical. It takes the same amount of delta-v to go from Earth's orbit to Mars's orbit, for example, regardless of direction. The one thing that changes this equation is aerobraking (or, when dealing with stuff whose structural integrity isn't important, lithobraking). Because drag works in one direction, that means that you can take advantage of it when arriving but not departing. For that reason, for example, it takes more delta-v to reach a transfer orbit to Mars from Earth than it takes to reach Earth from a transfer orbit to Mars. Technically that's not true, but when arriving at Earth, a great deal of delta-v can be provided by the atmosphere or the ground.
  - dhimes 13y ago
    I see that now. Thank you.
- dhimes 13y ago
  It's true tat it's been a long time since I played with this stuff, I work out for a circular orbit Total energy of an orbit is -G mp mS /(2 r) mp = mass of planet mS = mass of sun r = distance of planet from sun Even though this is a circular orbit, the basic result should be the same: the greater the distance from the sun the more energy the planet has, per unit mass. I think I understand your basic point, however, which is that whether we speed it up or slow it down we have to do something to the mass, and how much we do depends only on difference of the orbit radii. But note that adding 100 m/s or subtracting 100 m/s gives two different differences in r.

Yes it matters- the Sun is the main provider of gravity for this consideration.

by dhimes 13y ago

Yes it matters- the Sun is the main provider of gravity for this consideration.

Groxx 13y ago

orbit. gravity is canceled out by orbital velocity, as far as distance-from-sun is concerned. it costs Mars nothing to stay in orbit, and to go down to a -100m/s orbit or up to a +100m/s orbit costs the same amount of energy: (mass of mars) * (100m/s velocity change) if Mars and the Earth were not orbiting, I would completely agree. drop something from Mars and it'll land on Earth, and the reverse is not true. but they're not - drop something on Mars and it's just in Mars' orbit.

mikeash 13y ago
Delta-v requirements for transfers in space are generally symmetrical. It takes the same amount of delta-v to go from Earth's orbit to Mars's orbit, for example, regardless of direction. The one thing that changes this equation is aerobraking (or, when dealing with stuff whose structural integrity isn't important, lithobraking). Because drag works in one direction, that means that you can take advantage of it when arriving but not departing. For that reason, for example, it takes more delta-v to reach a transfer orbit to Mars from Earth than it takes to reach Earth from a transfer orbit to Mars. Technically that's not true, but when arriving at Earth, a great deal of delta-v can be provided by the atmosphere or the ground.
- dhimes 13y ago
  I see that now. Thank you.
dhimes 13y ago
It's true tat it's been a long time since I played with this stuff, I work out for a circular orbit Total energy of an orbit is -G mp mS /(2 r) mp = mass of planet mS = mass of sun r = distance of planet from sun Even though this is a circular orbit, the basic result should be the same: the greater the distance from the sun the more energy the planet has, per unit mass. I think I understand your basic point, however, which is that whether we speed it up or slow it down we have to do something to the mass, and how much we do depends only on difference of the orbit radii. But note that adding 100 m/s or subtracting 100 m/s gives two different differences in r.