# Can you jump off the counterweight of a space elevator at 1.2 x geostationary orbit? Is the assumption correct that an 80 kg person would weigh 2 kg?

Edit: This very silly question was answered in the comment by @pm-2ring - it's an orbit, you're weightless

Can you jump off the counterweight of a space elevator (https://en.wikipedia.org/wiki/Space_elevator) at 1.2 x geostationary orbit? Suppose an 80 kg person weighed 2 kg.

First, is kg-force estimate correct?

At 1.2 x geostationary orbit the object would have a velocity of circa 3600 m/s, 50500 km, so a 80 kg person would have a virtual weight of ~2 kg-force (https://www.thecalculator.co/others/Centrifugal-Force-Calculator-660.html).

This is equivalent to the person's weight on Charon. You would jump up and go back down - a smaller object, such as Deimos, is required to be able to jump off by one's own strength. https://journals.le.ac.uk/ojs1/index.php/jist/article/download/3092/2820 - the cutoff is Eros, where the Eros would weigh 60g. https://nssdc.gsfc.nasa.gov/planetary/text/eros.txt

I am assuming the answer is "no", an equivalent virtual weight of 2 kg is too large to jump off, and that object with virtual weight on a centrifuge act just as object do under real gravity.

• At the geostationary orbit, you're in orbit, so you're weightless. Commented Sep 8, 2021 at 15:31

I got your question.

First, let's point: You can't use these references about minimum mass/weight, because it's strongly dependent on the velocity. That article was made for man running about 45 kph.

But, as you've pointed in your question, the radius is 50500 km, and velocity, 3600 m/s.

Let's try a cleaner approach: analysis of Total Energy:

The Total Energy per kg of mass of the man, at one speed and radius, is:

$$\frac{E}{m}=\frac{v^{2}}{2}-\frac{GM}{R}$$

Your example doesn't meet the orbital conditions, but is in orbit only because the cable exerts a centripetal force at the counterweight. But, be adviced that without the cable, this counterweight (and the man on its surface) will search a new more external orbit.

So, let's return to Energy. Using your parameters in the above equation, we get:

$$\frac{E}{m}\approx-1.4\ MJ/kg$$

But, the minimum conditions to escape the orbit is to reach infinite radius at zero velocity, that implies in zero Total Energy. So if you want jump the counterweight and escape the Earth orbit, you need 1 MJ/kg amount of energy, that's not feasible for human being.

• So how 'heavy' (in kg force) would a body feel standing on on such counterweight compared to their normal weight? Commented Sep 9, 2021 at 14:20
• So, the effetive weight is given by EFF. WEIGHT = REAL WEIGHT - CENTRIP. FORCE (in the body frame). Divinding by REAL WEIGHT ($=\frac{GMm}{R^{2}}$), we have: $\frac{P_{eff}}{P_{real}}=\frac{R_{T}^2}{GMm}\left[\frac{GMm}{(R_{T}+h)^{2}}-\frac{mv^{2}}{R_{T}+h}\right]=R_{T}^2\left[\frac{1}{(R_{T}+h)^{2}}-\frac{v^{2}}{GM(R_{T}+h)}\right]$ , That yields $\frac{P_{eff}}{P_{real}}=\frac{R_{T}^2}{(R_{T}+h)^{2}}\left[1-\frac{v^{2}(R_{T}+h)}{GM}\right]$ Commented Sep 12, 2021 at 0:24
• ...and for a 80 kg body that results in __ ? Commented Sep 12, 2021 at 11:55

The elevator cable is exerting a centripetal force on the counterweight. If you exit through an airlock opposite the cable, you would drift away into a larger elliptical orbit.

• Exactly. And while standing on it, you would weigh how much compared to your habitual weight? Commented Sep 9, 2021 at 14:21