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I have a series of astronauts landing on an asteroid. I'd like to know the threshold of mass for them to not fly off into space. Specifications:

  • Assumption is all astronauts are under 180cm, 80kg.
  • Assumption is an astronaut may jump lightly and not fly off into space.
  • Request is a comfortable radius of said asteroid, which has an average density of 2 g/cm³ (pretty dense), with a mostly spheroid shape.
  • Weight and height includes equipment (this is far future); astronauts are fit.

In my story I'd like to see my astronauts be able to jump up into the 'sky,' but able to come back down without flying out forever.

If there are other factors I should include, please let me know. Also, I will accept a magnitude-of-order rather than an impossibly precise answer (thus not hard-science).

EDIT: AND I KNOW THIS AFFECTS THE ANSWER - 80kg, not 30kg. I will still accept the best answer if it used 30kg due to my error, but showed the math.

Mikey
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    No worry ur jumping astronauts will come back down... eventually unless external forces is applied. – user6760 Dec 17 '16 at 03:03
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    @user6760 the jump can exceed escape velocity. – JDługosz Dec 17 '16 at 09:18
  • @JDlugosz: yes you're right but the gravitational force is still there... unless of course external forces is applied. – user6760 Dec 17 '16 at 10:42
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    A ∆v applied at the surface that’s larger than the escape velocity will result in a hyperbolic course that escapes to infinity: it does not come back down, ever. I thought yiur first remark was confusing escape with forming an orbit (which requires a thrust that’s not against the surface). – JDługosz Dec 17 '16 at 11:50
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    30kg? Are they human? – Matt Bowyer Dec 18 '16 at 21:56
  • I can update my answer with your new numbers if you want. You'll have to specify how much of the 80kg is suit mass. Do we assume 70kg as a decent number for fit males? – user25972 Dec 20 '16 at 21:21
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    @user25972 - yes, that sounds suitable. That would be the threshold a 70kg person with 10kg suit mass. But I'll still accept an answer that used the 30kg, since that was my typing error, so long as the math is there. – Mikey Dec 20 '16 at 21:42
  • "EDIT: AND I KNOW THIS AFFECTS THE ANSWER - 80kg, not 30kg." Actually, the astronauts mass is irrelevant. – NPSF3000 Dec 22 '16 at 14:31
  • @NPSF3000 - how is it not relevant? I'm not being critical, just genuinely curious, because I assumed two masses were important in the equation of gravity and the amount of force needed to escape a planetary body. – Mikey Dec 23 '16 at 00:44
  • @Mikey Which falls faster? A 10-gram marble or a 10 ton stone (assuming no friction)? Same principle applies here. – NPSF3000 Dec 23 '16 at 06:25
  • @NPSF3000 - the formula for gravity: F = Gm1m2/r2, where F is the force due to gravity, between two masses (m1 and m2). Correct me if I'm wrong, but the mass of an object and the mass of a planetary body are important variables. (Not trying to be obtuse, just trying to figure this out) – Mikey Dec 23 '16 at 14:11
  • @Mikey I'm no expert... but that formula seems to scale force based on the second objects mass. A bowling ball that has a mass of 100x a marble has 100x the force. Consider however that f=ma... or for a=f/m. In other words, a much larger object requires a much larger force to have the same acceleration. Again, consider the marble vs the stone. – NPSF3000 Dec 25 '16 at 04:44
  • @NPSF3000 - yeah I'm familiar with a demonstration of a feather and a bowling ball in a vacuum as well. And also, I'm no expert at all. But the formula suggests that in extremely small bodies, there is a greater magnitude of order between the mass of two objects than small things vs Earth's mass. Again, not being argumentative - maybe someone with some physics background can help. – Mikey Dec 25 '16 at 20:26

2 Answers2

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How to calculate surface gravity

See here.

How much gravity do you want?

Lets say I can jump 1 meter on Earth (I can't). Then using kinematics equations, my initial velocity upwards was $$v_f^2 = v_i^2 + 2gd \rightarrow\sqrt{2gd}=4.4 \text{ m}/\text{s}.$$

I want my escape velocity to be comfortably above that, so lets say escape velocity is 10 m/s. You shouldn't be able to jump off that.

Escape velocity can be calculated as $$\Delta v = \sqrt{2gr}$$ where $r$ is the asteroid's radius and $g$ is surface gravity, which itself can be calculated from $$g = \frac{4}{3}\pi G\rho r$$ as shown in the above link. Setting escape velocity equal to 10 and plugging in we get $$10 = \sqrt{\frac{8}{3}\pi \left(6.67\times10^{-11}\right) (2000) r^2} \rightarrow r = 9459.$$

You cannot jump off a 10km radius asteroid. Probably.

**Edit*: Thanks to Mithrandir24601, I should have calculated mass as well. As he does in the comments: $$M = \frac{4}{3}\pi\rho r^3 = \frac{4}{3}(3.14)(2000)(9459)^3 \approx 7.1\times10^{15} \text{kg}$$

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kingledion
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  • I'd say you want to chose the escape velocity to be a good bit higher than 10m/s, as this is approximately the speed a human can achieve by muscle power (just sprinting horizontally, which would then put you into a low orbit). But excellent answer anyway. – Durandal Dec 18 '16 at 11:19
  • @Durandal I was thinking about that, but decided that with almost no gravity, you can't run that fast...your first step would take you way up and you can't take further steps for some time to build up any more speed. I any case, the best thing about math is that once you know the equations you can plug in your own assumptions :) – kingledion Dec 18 '16 at 19:52
  • Your first equation is incorrect. Since $\Delta V=at$ then $t=\frac{\Delta V}{a}$, next $\Delta s=\frac{at^2}{2}$ thus $\Delta V ^2= 2\Delta sa$, so your second is equation correct, but $\Delta V = V_f-V_i$ so setting $\Delta s=d$ and $a=g$ correct form of first equation would be $V_f^2-2V_fV_i+V_i^2=2gd$. Your first steps work because one of the values is 0, otherwise they wouldn't. $\Delta V^2=(V_f-V_i)^2 \neq V_f^2 - V_i^2$ – M i ech Dec 20 '16 at 22:09
  • To aid comparison, this gives the mass as $M = \frac{4}{3}\pi r^3\rho = \frac{4}{3}\pi (9459)^3 (2000) \approx 7.1 \times 10^{15} kg$ – Mithrandir24601 Dec 20 '16 at 22:11
  • @Miech Using my equation ($v_f^2-v_i^2 = 2ad$), with $v_f=1, v_i = 0, a=1$, d = 1/2. With $v_f=2, v_i=0, a=1$, d =3/2. That makes sense; if your initial speed is 1, you will travel 1 further distance unit while increasing speed 1 velocity unit at constant acceleration. Using your equation ($(v_f-v_i)^2 = 2ad$), for both the givens above, d = 1/2. That does not make sense. Here is a correct derivation of my kinematics equation. – kingledion Dec 21 '16 at 13:18
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    @kingledion Yeah, ok, I should stop doing math past midnight (so I shouldn't post this replay, actually). I should have used $\Delta s= V_it+ \frac{at^2}{2}$ after inserting $t=\frac{\Delta V}{a}$ it gives $a=V_i\frac{\Delta V}{a}+\frac{\Delta V^2}{2a}$ and thus $2as=2V_iV_f-2V_i^2+V_f^2-2V_iV_f+V_i^2=V_f^2-V_i^2$. So I concede I was wrong, I still hate the method you linked, geometric methods are awful. – M i ech Dec 21 '16 at 23:09
  • you can add resulting gravity and height of the possible jump and maybe the time of that jump (how long it will last) - nice answer – MolbOrg Dec 22 '16 at 02:46
  • @Miech Unfortunately, geometrically is how they teach it in high school, and I never learned me no better in college. – kingledion Dec 22 '16 at 13:46
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The maximum height that a man can jump is a bit over 70cm for the very best athletes. This is because the only thing that matters is how high your centre of gravity rises. I will assume 80cm.

Work is force times distance. Weight is the force $W = gm_0 $ and when you jump a height $h$ under Earth gravity of 9.8m/s with a weight $m_0$, $E = hgm_0$ is the work that gravity did to bring you to a stop at your maximum height (all of the kinetic energy being converted to potential energy).

The work you performed to initiate the jump is also force times distance. It equals the gravitational potential energy at the top of the jump exactly. When jumping on an asteroid, the length of your legs and your strength we will assume don't change (a fabric-based space suit has a springiness in the legs, so this is an approximation). You perform an equal amount of work to what you did on Earth, namely $E = 0.8m \times 9.8 m/s^2 \times m_0$.

Escape velocity is given by (see Space Mission Engineering: The New SMAD edited by Wertz, Everett & Puschell, 2011, p. 201) $V_e = \sqrt{2GM/R}$ where $G = 6.674 \times 10^{-11}m^3 kg^{-1}s^{-2}$ is the gravitational constant, M the mass of the central body and R your distance from its centre (I assume for simplicity it is circular).

Density $\rho$ is in the region of $1000kg/m^3$ for icy bodies and $3000kg/m^3$ for rocky ones. In the case of a nickel-iron body we will approximate the overall density as $8000kg/m^3$.

We want to calculate the diameter R of a body where your kinetic energy at escape velocity equals $E$ above.

$$0.8m \times 9.8m/s \times m_0 = E = KE = 0.5m_0 \times V_e^2$$ $$0.8m \times 9.8m/s^2 = 0.5 \times V_e^2 $$ $$ 15.68 m^2/s^2 = V_e^2 = 2GM/R $$ still assuming a sphere $$ M = \rho \times V = \rho \times 4/3 \pi R^3 $$ $$15.68 m^2/s^2 = 8/3 G \rho \pi R^2 $$ $$ \sqrt{(2.8 \times 10^{10} / \rho) m^{-1} kg } = R$$

Substituting the densities from above, we get roughly $R_{icy} \approx 5300m$, $R_{rocky} \approx 3100m$, $R_{iron} \approx 1900m$.

Now let's see what happens with your numbers. The astronaut can still do an equal amount of work to jump but their mass will be $m_1 = m_0 + m_{suit}$ on the asteroid, $\rho = 2000kg/m^3$. A NASA study from 1996 (see The Origins and Technology of the Advanced Extravehicular Space Suit by G.L. Harris, American Astronautical Society History Series Volume 24, 2001, p. 455) specified a maximum pressure suit assembly mass of 27kg for missions to Mars. This excludes the life support systems, for comparison Apollo had (see p. 440) 63.2kg of life support on a suit of 100kg total. We'll assume the future suit is a form-fitting elastic suit with almost all mass in life support, giving no springiness in the legs and massing just $m_{suit} = 30kg$ total. Note: You gave the astronaut weight as 30kg. This is a small child so I assume 70kg is intended instead.

$$ 0.8m \times 9.8m/s^2 \times m_0 = E = KE = 0.5 m_1 \times V_e^2 $$ $$ 0.8m \times 9.8m/s^2 \times 70kg = 0.5 \times 100kg \times V_e^2 $$ $$ 10.976 m^2/s^2 = V_e^2 = 2GM/R = 8/3 G \rho \pi R^2 = 8/3 \times 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2} \times 2000 kg/m^3 \times \pi \times R^2$$ $$R^2 = 9.815 \times 10^6 m^2 $$ $$R \approx 3100 m $$

It only happens to be close to the rocky body approximation above because the increase in mass was made up by the decrease in gravity.

Edit: Finally, since you want mass: $$ M = 4/3 \pi R^3 \times \rho = 4/3 \times 2000kg/m^3 \times \pi \times (3100m)^3 = 2.50 \times 10^{14} kg $$

Really finally this time: Since you updated the mass to 80kg and we may assume a suit mass of 10kg included in that, here's how the answer would change from my 100kg with 30kg suit mass answer:

$$V_e^2 = 10.976m^s/s^2 \times 100kg/80kg = 13.720m^2/s^2$$ $$R \approx 3100m \times \sqrt{100kg/80kg} \approx 3500m$$ $$M = 2.50 \times 10^{14}kg \times (\sqrt{100/80})^3 = 3.6 \times 10^{14}kg$$

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