The acceleration a ballistic projectile would experience at a certain height above a planet (neglecting any atmosphere) is given by

$$a = -\frac{GM}{(R + h)^2}$$


  • $G$ is the gravitational constant $6.67 \times 10^{-11}$ m3s-2kg-1
  • $h$ is the altitude above the planet's surface in meters
  • $M$ is the planet's mass in kg
  • $R$ is the planet's radius in meters

I need an expression with which I can get current height $h$, time and remaining velocity $v_r$ through the initial velocity $v_i$ the projectile starts with and the variables listed above.

  • 3
    $\begingroup$ looks like a fairly simple calculus problem to me $\endgroup$ – JCRM Feb 5 at 8:44
  • $\begingroup$ @JCRM exactly. A link might do too, if you don't think it's worth your time. $\endgroup$ – justthisonequestion Feb 5 at 8:54
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    $\begingroup$ Where did the part of the question about the influence of other planets come from? That massively complicates the issue, and seems to change the whole point of the question. $\endgroup$ – Organic Marble Feb 5 at 14:55
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    $\begingroup$ I nearly rolled it back when I saw it @OrganicMarble -- but space.meta.stackexchange.com/questions/1419 $\endgroup$ – JCRM Feb 5 at 17:54
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    $\begingroup$ @OrganicMarble I think I see what happened. The title says "effect of a planets" (and not planet's with an apostrophe) and I mis-read it as "effect of planets", missing the "a". Luckily I don't write software for Boeing (but maybe it wouldn't hurt, considering their batting average?) $\endgroup$ – uhoh Feb 10 at 12:23

Here are some observations you can make, before attempting any calculus with respect to time.

First, there's the potential energy of the gravity well of the planet:

$$E_{potential} = -\frac{GMm}{r}$$

Thus, if the projectile has enough kinetic energy, it can escape:

$$E_{kinetic} + E_{potential} > 0$$

$$\frac{v_i^2m}{2} -\frac{GMm}{R+h_i} > 0$$

$$\frac{v_i^2}{2} -\frac{GM}{R+h_i} > 0$$

Otherwise, it will fall back. In that case, it will reach a certain maximulm altitude $h_{max}$, which we can calculate since it will have zero kinetic energy there:

$$\frac{v_i^2}{2} -\frac{GM}{R+h_i} = -\frac{GM}{R+h_{max}}$$ $$(R+h_{max})\left(\frac{v_i^2}{2} -\frac{GM}{R+h_i}\right) = -GM$$ $$h_{max}\left(\frac{v_i^2}{2} -\frac{GM}{R+h_i}\right) = -GM - R\left(\frac{v_i^2}{2} -\frac{GM}{R+h_i}\right)$$ $$h_{max} = -\frac{GM}{\frac{v_i^2}{2} -\frac{GM}{R+h_i}} - R$$

In the case id does escape, the velocity will decrease to a certain $v_{\infty}$:

$$v_{\infty} =\sqrt{v_i^2 -\frac{2GM}{R+h_i}}$$

| improve this answer | |
  • $\begingroup$ thanks. I think that's as close to a solution one can get in this case. $\endgroup$ – justthisonequestion Feb 26 at 13:41

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