Also, you need exactly as much energy to get back to earth as you did to get to the moon. On the way up you need a booster to get above the atmosphere. Then on reentry, you use the atmosphere to bleed off all that energy (much smaller spacecraft on the return trip). Reentry does the same amout of delta-v, pound for pound, that the entire booster stack does on ascent to TLI (roughly).
Moon gravity is about 1/6 that of earth’s. So it’ll need less than 1/6 the force to escape moon’s gravity. Less because the diameter of the moon is smaller therefore less time under power to escape the smaller atmosphere.
The first part you said is true. But in orbital mechanics it's all about delta-v or change in velocity. So the second part is incorrect. You're still coming back to earth at 24,500 mph and need to bleed off that energy.
Also, you need exactly as much energy to get back to earth as you did to get to the moon. On the way up you need a booster to get above the atmosphere. Then on reentry, you use the atmosphere to bleed off all that energy (much smaller spacecraft on the return trip). Reentry does the same amout of delta-v, pound for pound, that the entire booster stack does on ascent to TLI (roughly).
Moon gravity is about 1/6 that of earth’s. So it’ll need less than 1/6 the force to escape moon’s gravity. Less because the diameter of the moon is smaller therefore less time under power to escape the smaller atmosphere.
The first part you said is true. But in orbital mechanics it's all about delta-v or change in velocity. So the second part is incorrect. You're still coming back to earth at 24,500 mph and need to bleed off that energy.
I was just talking of the escape from surface/gravity part. Hadn’t thought of the braking aspect.