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Does it require more energy for a space vehicle to leave low-earth orbit, or the surface of the moon?

I’m in a debate over which would be the cheaper, more efficient platform for launching a mission to Mars, a space station or a moon base. Regardless of all the other various advantages and disadvantages of these two options, which one would require the least amount of energy for a vehicle to leave? I don’t know enough physics to figure out the energy required to leave a low-earth orbit vs. the surface of the moon.

4 COMMENTS

  1. The moon.
    If an object is already in orbit, you can, for all intensive porpoises, consider it to be in zero gravity. You would not need any energy to leave Earth’s gravity well, only the energy necessary to accelerate the ship from a stationary position.
    Addendum: starrysky – “Alexis is not correct about “zero gravity” in earth orbit. Objects in spaceships or stations around earth are “weightless” in relation to their vehicle or container. But both need a lot of power to continue to climb farther out of the “gravity well” of the earth, and of the earth-moon system to get to other planets.”
    I’m sorry, but this just isn’t correct.
    If you stand on the surface of the moon and throw something (say, a wrench) directly away from it, it will travel a distance, slow, stop, and return to the surface.
    If you throw a wrench directly away from the Earth from an object in orbit, it will continue to circle the Earth, spiraling away and receding at a constant velocity. It will *never* return to Earth.
    The very act of putting something into orbit entails imparting it with enough energy to exactly equal the escape velocity of the object it is orbiting. That is what an orbit *is*, and why something in orbit maintains its position. Any *further* energy applied to the object will cause it to recede from the object it is orbiting, forever.

  2. The guys at NASA are planning to leave from earth orbit.
    Not sure if this is because of safety concerns, or if it is less power. I believe both to be true.
    It takes a lot of energy to leave earth orbit, go to moon, decelerate, land on moon using power, and then depart the moon’s surface for somewhere besides earth. It seems a lot easier and cheaper to depart from earth.
    Alexis is not correct about “zero gravity” in earth orbit. Objects in spaceships or stations around earth are “weightless” in relation to their vehicle or container. But both need a lot of power to continue to climb farther out of the “gravity well” of the earth, and of the earth-moon system to get to other planets. And “intensive porpoises” or any other cetacean or fish is not relevent.
    Where an advantage might come is if there was a launch system built on the moon. A “mass driver” or rail gun could propel lots of matter from the moon without rockets. A solar-electric powered electromagnetic acceleration track could catapult supplies from the moon. Oxygen and other gases, perhaps made from ice below the surface, could be compressed and frozen into packages that would be shot into interplanetary space. These would refuel rockets on way to, around, or back from Mars or other planets.

  3. You want to find the delta-vee required for a minimum energy escape trajectory.
    The gravitational constant.
    G = 6.67428e-11 m³ kg⁻¹ sec⁻²
    Moon.
    M = 7.3477e22 kg
    R = 1.7371e6 m
    Escape speed from surface = sqrt(2GM/R) = delta vee
    Escape speed from surface = 2376.2 m/s
    Earth.
    M = 5.9736e24 kg
    R = 6.3781e6 m
    H = 1.0e5 m
    Circular orbital speed at 100 km altitude = sqrt[GM/(R+H)]
    Circular orbital speed at 100 km altitude = 7845.1 m/s
    Escape speed at 100 km altitude = sqrt[2GM/(R+H)]
    Escape speed at 100 km altitude = 11094.6 m/s
    Delta-vee needed to escape from orbit = 3249.5 m/s
    It’s easier to escape from the moon’s surface than it is to escape from low Earth orbit.
    It requires more energy to escape from low Earth orbit than it does to escape from the surface of the moon.
    Alexis is not correct when she says:
    “The very act of putting something into orbit entails imparting it with enough energy to exactly equal the escape velocity of the object it is orbiting. That is what an orbit *is*, and why something in orbit maintains its position. Any *further* energy applied to the object will cause it to recede from the object it is orbiting, forever.”
    Boosting an object into a circular orbit requires half the energy needed for a minimum escape orbit because the escape speed is equal to the square root of 2 times the circular orbit speed.
    In between circular orbits and the minimum energy escape trajectory is the range of possible elliptical orbits, each having an energy greater than that of a low circular orbit but less than that required to escape from the planet.
    escape speed = sqrt( 2GM / r )
    circular orbit speed = sqrt ( GM / r )
    elliptical orbit speed = sqrt [ GM (2/r − 1/a) ]
    where…
    a = the semimajor axis of the ellipse
    r = is the current distance of the spaceship from the planet
    If you’re in a circular orbit and add only a little more energy, you will be in an elliptical orbit with perigee located at the position where you fired your rockets. In order to escape from the planet from a circular orbit, you would have to add a SUFFICIENT amount of energy, to increase your speed relative to the planet by about 42%.

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