# How does the gravity of Earth change as you climb a space elevator?

I'm doing research for a movie. If you were in a space elevator/tower at an altitude of 250 miles, how much gravity would you experience? (I've heard various estimates of .9Gz and .65Gz)

How about if you were at an altitude of 22,236 miles?

• Commented Apr 7, 2016 at 18:23
• G is usually used for gravitational constant. g is used for surface gravity (on earth around 9.8 m/sec^2). I know it's not kosher to use an apostrophe for plural but usually write more than 1 g as several g's. Commented Apr 7, 2016 at 22:44

Organic Marble gives a good answer. Hohmannfan gives a better answer if you want the net acceleration someone at that point of the elevator would feel.

Using both gravity and the so called centrifugal force, someone would feel .88 g's 250 miles up the elevator.

At 22237 miles altitude I get -7.23 millionths of a g which is darn near zero. Just a slight tug upward so that altitude is just a tad higher than geosynchronous orbit.

Here is a spreadsheet NetAccelOnElevator.xls. User inputs altitude in miles to the colored cell at the top. Below will appear the net acceleration in g's.

If you want to consider the centripetal force too, you can do that by just adding a little more to the equation already suggested by Organic Marble:

$$g_h=g_0 \left(\frac{r_e}{r_e + h}\right)^2 - \frac{4\pi^2 (r_e+h)}{t^2}$$

Where $h$ = height above sea level, $r_e$ = radius of Earth, $g_0$ = sea level gravity and the additional $t$ is the length of the Earth's rotational period.

The acceleration profile looks like this:

For the two values you have chosen, the acceleration is 0.88g for 250 miles, and 0g for 22236 miles, usually known as the geostationary altitude,

• Minor detail: t would be about a minute shorter than 24 hours because the Earth is moving around the sun as it rotates: en.wikipedia.org/wiki/… Commented Apr 7, 2016 at 19:18
• @AndrewW. Fixed now. Next thing is to adjust for the gravitational influence of the Moon :P Commented Apr 7, 2016 at 19:24
• @Hohmannfan Why not adjust for the gravitational influence of J002E3 while you're at it? :-)
– user
Commented Apr 7, 2016 at 20:12
• Thanks for the answers. I really appreciate it. However, I neglected to note (my bad) that the reason I was asking is my algebra skills have atrophied over the years and I keep getting whacky answers when I did the calculations myself. I was hoping someone could do the calculations for me. Commented Apr 7, 2016 at 22:00
• @NealWiser Added to the answer now, together with a graph to visualize it. Commented Apr 7, 2016 at 22:17

I could calculate it for you, but using the "Teach a Man to Fish" principle, here's a simple equation (lifted from Wikipedia)

$$g_h = g_0 \left( \frac{r_e}{r_e + h} \right) ^2$$

where:

$h$ = height above sea level

$r_e$ = radius of Earth

$g_0$ = sea level gravity

Use the units of your choice.

Note: this accounts for altitude effects only and not centripetal force.