How far above the Earth's surface would you have to go to lose half your weight?

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Answer 1

This is an easy problem. We're going to ignore everything but the earth and you.

Fg = GMm/r² = mg

From this defintion, we see that GM/r² = g.

To cut your weight in half, you want to cut g in half.

So..

6.67 * 10^-11 * 6.0 * 10^24 / r² = 9.8 / 2

4.002 * 10^14 / r² = 4.9

Solving for r

r = 9.04 * 10^6 m, or about 9040 km above the earth's center or about 9040 - 6400 = 2640 km above the earth's surface.




Tito's explanation makes perfect sense. He remembered that the force of gravity is proportional to 1/r², so he did the appropriate math steps to figure it out. Kudos to him :)

Answer 2

If I recall correctly, your weight (attraction of gravity for your mass) depends upon your distance from the center of mass of the planet or moon and varies inversely as the square of that distance. On the Earth's surface, you are about 4000 miles from its center. If you doubled that distance, i.e., 4000 miles out in space, your weight would only be 1/4 of what it is now. To get 1/2 of your weight as the result, therefore, you want 4000 times the square root of two (about 1.414), and that would come to about 5656 miles, or 1,656 miles above the surface. I can't guarantee that result, it's just a guess based on what I remember from grade 12 physics in 1956.

Answer 3

nicely, once you're in touch in what the size says, you should in simple terms flow to the Moon - you should promptly lose 5/6 of your weight as stated with assistance from that pesky bathing room device. Too undesirable your mass doesn't definitely change, even if, so in case you had to seem in good structure and trim, i'd advise the community living house of mirrors. locate one it truly is convex alongside the vertical axis. in case you won't be able to make each of the girls in person-friendly words seem at your moon scale, equator scale, or your mirrored image in the slimm-o-matic reflect, then I advise the eating routine. Now in case you've been to flow to the equator walking, that can make your decrease weight measured a lot more advantageous everlasting. ;)

Answer 4

At the earth's surface of average radius Ro, GMe/Ro^2 = g

What you need is at height h above the surface of the earch, (distance = Ro+h from center), where the g' = GMe/(Ro+h)^2 = 1/2 g

So 2Ro^2 = (Ro+h)^2

Take square root on both side

1.41 Ro = Ro + h
So h = 0.41 Ro

Answer 5

It would have to be far enough so that the acceleration of gravity is 4.9 m/s². I don't believe there is an exact way to find how far that would be. I could be wrong though.