At what rate does the earths gravity lessen as you leave the earths surface?

Question

For example; A 200 lb man standing on the ground, how much would he weigh traveling on an airplane at 30,000 ft? At what point would he start to weigh less?

Answer 1

it lessens with 1/(r^2)
but at the surface of earth, r is a few thousand km (it's measured from the core). so r+30,000 feet is almost the same as r
so he'd be about 200lb

Answer 2

W = mg = GmM/R^2; where m is the man's mass, g ~ 32.2 ft/sec^2, G is a constant, M is Earth's mass, and R (~ 4,000 mi) is Earth's radius (i.e., center to the surface). [See source.]

w = mg' = GmM/(R + h)^2; where g' is the acceleration due to gravity at R + h distance from Earth's center and h = 30,000 ft ~ 6 mi.

Then the ratio w/W = g'/g = (R/(R + h))^2 is true; so that g' = g[R/(R + h)]^2 will give you the new g' at altitude h above Earth's radius R. In other words, you 200 lb man would start to weigh less as soon as R + h > R or as soon as he lifts off with h > 0.

For example, assume a satellite is orbiting at h = 2R, two Earth's radii above the surface. Then g' = g [R/(R + 2R)]^2 = g(1/3)^2 = g/9; so the satellite's acceleration due to gravity at h = 2R is 1/9 what it would be on the surface of Earth.

One more example, assume h = 6/4000 R ~ 1.5 X 10^-3 R. Then g' = g(R/(1.0015R))^2 = g(1/(1.0015))^2 ~ g. So your 200 lb passenger will weigh very slightly less than 200 lb at 30,000 ft because g' ~ g at that altitude.

Of course R <> 4000 everywhere; its bigger across the equator than across the poles. Mother Earth is a bit dumping; she's actually an oblate spheroid....meaning plump around the middle.

Answer 3

It lessens by the square of the distance from the centre of the Earth. A 200 pound man at 30,000 feet would weigh 199.32 pounds. He begins to weigh less as soon he leaves the surface of the Earth. A man on top of Mount Everest weighs less than he does on the equator.

Answer 4

It falls off exponentially as you progression far off from the source of the gravity: tension of gravity = G * m1 * m2 / r^2 r^2 is the significant element... it fairly is the gap between the two gadgets (to that end you and the earth). the interior the Earth might have extra gravity than the exterior (especially because of the fact many of the mass of the Earth is on the middle). So definite, the nearer you're to the middle of the Earth the extra beneficial the gravity and it falls of further and further as you pass away the Earth.

Answer 5

The earth's gravity is unchanged by it's distance from you.

What changes is the force Earth's gravity exerts upon you. This can be calculated by equation F=G*m1* m2 / r^2

F is the gravitational force
G is the gravitational constant (check a physics reference)
m1 is the mass of the first body (the Earth)
m2 is the mass of the second body (you)
r is the distance between the two bodies, in this case you and the center of the Earth.

For your example, 30,000 feet is a negligible distance given that the Earth's radius is around 600,000 km (if I remember correctly), so his weight would be almost the same. However, if you look at the equation, you'll see that the man will weigh less for every micron he travels away from the center of the planet.

Answer 6

The force of gravity is F(g) = G * m1 * m2 / r². Since the mass of the object, Earth, and G are all constants, you only need worry around r (distance from the center of the Earth).

The radius of the Earth is varies from 6,356.750 km — 6,378.135 km (it is not a perfect sphere and bulges slighly at the equator). Assume 6,360 km. 30,000ft is 9.15km

F(g, on Earth) = constant * (6,360)*(6,360)
F(g, at +9.15km) = constant * (6,360+9.15)*(6,360+9.15)

F(g, at +9.15km) = F(g, on Earth) * (6360*6360) / (6369.15*6369.15)

F(g, at +9.15km) = 0.9971 * F(g, on Earth)

The force of gravity at +9.15km is about 99.71% that on the surface below. A 200lb man would weigh 199.41 lbs at 30,000ft.

However, you could also apply the calculation over the equator or poles and get slightly different results.

Finally, the surface of the Earth is not smooth. Land rises up and oceans fall to many tens of thousands of feet. Local density differences in the planet would likely make the above 0.29% difference hard to detect. For example, if you are flying 30,000ft above sea level and there is a 20,000ft mountain below you, vs. flying at 30,000ft over a deep ocean. Local height and density of the surface below you is likely going to make it hard to notice the difference in the man's weight.