If you could reverse gravity, to make it repulsive instead of attractive, what velocity would you achieve on leaving Earth? A simple question really, but I have completely forgot how to calculate this sort of thing.

Also after leaving Earth, if you could reverse the Sun's gravity (on yourself) what velocity could you achieve leaving the solar system?

  • $\begingroup$ What "reverse" means here? $\endgroup$ – unsym Feb 2 '13 at 16:43
  • 6
    $\begingroup$ Equate gravitational potential energy at r = R_earth with kinetic energy and you're done. :) $\endgroup$ – Robert Filter Feb 2 '13 at 16:44
  • 1
    $\begingroup$ @RobertFilter This is the answer, you should write it up and give an explanation. The question is rich. It has conservation of energy, and touches on + and - charges in gravitation. $\endgroup$ – kηives Feb 2 '13 at 17:28
  • $\begingroup$ @kηives I don't know, I think for such a straight forward question I prefer just to help the questioner to get the answer her/himself :) $\endgroup$ – Robert Filter Feb 4 '13 at 7:45
  • $\begingroup$ @RobertFilter it's not that straightforward, because you have to integrate to infinity to find the gravitational potential - see my answer. $\endgroup$ – Nathaniel Feb 5 '13 at 12:57

The acceleration due to gravity is approximately $9.8\ ms^{-2}$ at the surface and falls off as you rise with the inverse square law.

So you would start accelerating upwards at $9.8\ ms^{-2}$ but as you rose upwards your acceleration would drop. It would never actually hit zero (Xeno's paradox) but would get vanishing close. At this height you would carry one moving away from the earth at a constant velocity.

It's generally accepted that you've entered "outer space" when you pass the Kármán line at 100 km. At this height the acceleration due to gravity is approximately $9.515\ ms^{-2}$. Taking the average value (to make the calculation simpler):

  • Acceleration $= 9.66\ ms^{-2}$
  • Distance = $100\ km = 100,000\ m$


$v = \sqrt{2da}$ (from here)

Which gives us:

$v = \sqrt{2 \times 100,000 \times 9.66} = 1,389.96\ m/s = 5,003,870.50\ km/h$

You would however, be still accelerating.

  • $\begingroup$ After a while, this approach will no longer be valid, as your velocity becomes more and more relativistic. This can be fixed by explicitly applying conservation of energy with the relativistic formulas. $\endgroup$ – Wouter Feb 3 '13 at 16:35
  • $\begingroup$ 1. It's Zeno's paradox, not Xeno. 2. This has nothing to do with Zeno's paradox. You might 'never hit infinity' if you assume that $F(r)=\frac{GMm}{r^2}$ for all $r$. I do not know if you would actually 'never hit zero' if you consider things like string theory, quantum energy, relativity etc. 3. What does this have to do with never hit zero? Since we always have acceleration in the direction of motion, what we're looking for the limit of the velocity, $\lim_{t\to\infty}v=v_{max}$ $\endgroup$ – raindrop Feb 3 '13 at 17:44
  • $\begingroup$ Nathaniel's answer gives us the maximum velocity, but again this has nothing to do with Zeno's paradox. Zeno's paradoxes are , while Physics is real $\endgroup$ – raindrop Feb 3 '13 at 17:57
  • $\begingroup$ @Raindrop - I always get that wrong. I blame Douglas Hofstadter (or Terry Pratchett). I just used it as an example of something asymptotically approaching a value but never reaching it. $\endgroup$ – ChrisF Feb 3 '13 at 22:27
  • $\begingroup$ You're right! Zeno's paradox has a limit $\lim_{n\to\infty}\sum_{i=2}^{n}2^{-n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... $ and the answer to your question is $v_{max}=\lim_{t\to\infty}v$ Both are examples of $\lim_{something\to\infty}$. $\endgroup$ – raindrop Feb 4 '13 at 0:20

A straightforward but complicated way to do this would be to note that, at any given time $t$, your acceleration will be given by $$ \frac{d^2h}{dt^2} = \frac{GM}{h^2}, $$ where $h$ is your height above the centre of the Earth. $G\approx 6.7\times 10^{-11}\,\mathrm{m^3kg^{-1}s^{-2}}$ is the gravitational constant and $M\approx 6.0\times 10^{24}\,\mathrm{kg}$ is the mass of the Earth. This equation comes from Newton's $F=ma$, and then cancelling $m$ from both sides. You could then solve this differential equation with the initial condition that, at time $t=0$, $dt/dh=0$ and $h=r_E$ (the radius of Earth, $\approx 6.4\times10^6\,\mathrm{m}$) to get $h(t)$, and then take the limit of $dt/dh$ as $t\to\infty$. This would give you the asymptotic speed that you will approach once you are far enough away from the Earth that its gravity is insignificant.

However, a simpler way is to note that whenever you move upwards by a small distance $\delta h$, your kinetic energy must increase by $mg\delta h$. (Normally this would be a decrease, of course.) Here $g$ is the acceleration due to gravity, which is given by $GM/h^2$. Since your initial kinetic energy is 0, your total kinetic energy after travelling from height $h_0 = r_E$ to height $H$ is given by $$ m\int _{h_0}^H \frac{MG}{h^2}dh = m\left[ -\frac{GM}{h} \right]_{h_0}^H = GMm\left(\frac{1}{h_0} - \frac{1}{H}\right). $$ (Note that we're now integrating with respect to $h$ rather than $t$, which makes it much easier.) Taking the limit as $H\to\infty$ gives $\frac{GMm}{h_0}$. This is your eventual kinetic energy, which gives us $$ \frac{1}{2}mv^2 = \frac{GMm}{h_0}, $$ or $$ v = \sqrt{\frac{2GM}{h_0}}. $$ Plugging in the numbers (with M the mass of Earth and $h_0$ its radius), we get $v=11.8\,\mathrm{kms^{-1}}$, which again is an asymptotic speed that you will tend towards but never quite reach.

Of course, the above assumes no influence from the Sun or any other object. However, using this method we can calculate the maximum speed you can reach by noting that your final kinetic energy must be the sum of the potential energy you've gained from the Earth's inverted gravity field, plus the energy you've gained from the Sun's inverted gravity field. This gives us $$ \frac{1}{2}mv^2 = \frac{GM_\text{Earth}m}{r_E} + \frac{GM_\text{Sun}m}{r_O} \,\,, $$ (where $r_O$ is the radius of Earth's orbit), or $$ v = \sqrt{2G\left(\frac{M_\text{Earth}}{r_E} + \frac{M_\text{Sun}}{r_O}\right)}\,\,. $$ Plugging in the numbers again (Wolfram Alpha is incredibly useful for this sort of thing. It even checks the units for you) we get $v=43.87\,\mathrm{km\,s^{-1}}$ for the speed you would approach after leaving the Solar system.

  • $\begingroup$ +1 for mentioning the Sun. I wonder if we need to compensate for Jupiter? $\endgroup$ – Jitter Jan 10 '14 at 7:30

Imagine it the other way round. The velocity at which you hit the Earth's surface after falling from the sky is the same velocity at which you would reach the sky after 'falling reverse gravity style' from the Earth's surface.

The velocity you would reach after falling from outside the solar system to reach some 'outside Earth' point is the same velocity you would reach after 'falling reverse gravity style' from that 'outside Earth' point to outside the solar system.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.