A view of comet 67P as seen from the Rosetta spacecraft ESA/Rosetta/NAVCAM

I recently looked at the physics of an astronaut jumping off the Rosetta spacecraft as it passes comet 67P. Based on my calculations, yes -- an astronaut could probably jump in a way that he or she would land on the surface of the comet (if the spacecraft had a low enough relative velocity).

Could an astronaut even stand on a comet? I don't know. I left that as a homework question for you. Maybe you're still working on the homework.


Ok, so let's say that you can indeed stand on the surface of a comet. Could you walk? I'm going to go ahead and guess "no". Why not? Two reasons. First, the gravitational field is quite low such that you probably wouldn't be able to always keep one foot on the ground during the motion (speed walking definition). Second, an astronaut would probably be wearing a space suit. With a space suit, it can be difficult to bend legs and arms.

Here is an example of what I mean. This is a video of astronauts "walking" during the Apollo 17 mission.

Read next Comet 67P gives up more of its secrets: Nasa spots a landslide on Churyumov–Gerasimenko Comet 67P gives up more of its secrets: Nasa spots a landslide on Churyumov–Gerasimenko

In order to see this embed, you must give consent to Social Media cookies. Open my cookie preferences.

Ok, so this is kind of like skipping. Let me just ask "can you move on the surface of a comet"?


Assumptions

Without actually going to a comet's surface and testing out some stuff, I am going to have to make some assumptions.

Comet 67P is spherical and locally flat. This means that I can estimate the gravitational field on the surface as some constant value. Also, when an astronaut jumps, the starting and landing height will be the same.

Comet 67P has a radius of 2km with a mass of 3.14 x 10<sup>12</sup> kg.

When an astronaut "skips", the astronaut is just like projectile motion with a constant vertical acceleration and a zero horizontal acceleration.

The astronaut can push horizontally on the surface of the comet. The coefficient of static friction between a space boot and the surface is 0.9 (I just picked something super high - maybe there are spikes on the boots).

When an 80 kg astronaut jumps, he or she can use 376 Joules in this motion ( the same as when jumping off a spacecraft).

The rotational effects of the comet (since it is spinning) will be ignored (for now).

That should be enough to get us started.

Read next Hubble spots 'big brother' of Halley’s Comet being torn apart by white dwarf Hubble spots 'big brother' of Halley’s Comet being torn apart by white dwarf

Jumping on the surface


What if an astronaut jumped straight up? How high would this astronaut rise before falling back down? This isn't too difficult.

Let me start by assuming that the astronaut doesn't go high enough such that the gravitational field is mostly constant. What is the gravitational field on the surface of this comet? Since I have assumed a spherical comet, I can use the gravitational force for a point mass:

Here G is the gravitational constant and MC is the mass of the comet. Putting in these values, I get a gravitational field strength of 5.24 x 10<sup>-5</sup>N/kg.

Read next How did life on Earth begin? Crashing comets may hold the key to unravelling the mystery How did life on Earth begin? Crashing comets may hold the key to unravelling the mystery

But how high would the astronaut go? Since we are looking at a change in position, this would be a straight forward application of the work-energy principle. This says that the work done on some system is equal to the change in energy of that system. If I include the astronaut plus the comet as the system, then there would be nothing left to do work on this system (so zero work).

Then I could write:

already know the change in energy of the human. The human uses 376 Joules, then the human will decrease in energy by 376 Joules. I also know the change in kinetic energy. Since the human (astronaut) starts jumping from at rest and ends at the highest point (also at rest), the change in kinetic energy is zero Joules.

If the gravitational field is constant, then the change in gravitational potential energy would be the following (top) and then by putting this all together, I can solve for the change in y for this jumping astronaut (bottom):

Read next Mission complete! Rosetta successfully 'crashes' into comet 67P Mission complete! Rosetta successfully 'crashes' into comet 67P

Ok, we have a problem. If the astronaut jumps 90 km high it's probably not a good idea to assume a constant gravitational field.

If the gravitational isn't constant over this displacement, then I need to use a better model for the gravitational potential energy.

In this expression, G is still the gravitational constant. The r's are the distances from the centre of the comet both at the beginning and ending of the motion. The value for r1 would be the radius of the comet (2,000 m). What if I want to calculate the amount of energy to get the astronaut an infinite distance from the comet? For this, the final position would approach infinity such that one over that distance would approach zero. This energy would be:

Read next Rosetta spots the lifeless body of its long-lost Philae lander on Comet 67P Rosetta spots the lifeless body of its long-lost Philae lander on Comet 67P

In case it's not clear, 8.38 Joules is LESS THAN 376 Joules. If the astronaut jumped and used 376 Joules, this would be more than enough energy to jump so high that the astronaut would never return -- ever. Well, at least not return due only to that gravitational interaction.

Escape velocity

If an object at the surface of a gravitational object has a large enough speed, the object can move away from that object such that it will never "fall back" to the surface. The speed at which an object can "escape" is called the escape velocity.

Suppose an object starts at the surface of the comet with a velocity of v1. It then moves away from the comet until it is very, very far away with a velocity of zero. If the object and the comet is the system, then there is no work done and the work-energy equation becomes:

Read next Nasa witnesses ancient comet being vaporised by the Sun Nasa witnesses ancient comet being vaporised by the Sun

Since the gravitational potential is zero at a very great distance and the kinetic energy is also zero, both of the "2″ terms will be zero in the above equation. If I put in an expression for the kinetic energy and the gravitational potential energy, I can solve for the initial velocity.

If I put in values for the radius and mass of comet 67P, I get an escape velocity of 0.46 m/s. Yup. This isn't Planet of the Apes. It doesn't take much to escape from this comet. A human astronaut should easily be able to jump off and never come back.

Does the comet's rotation even matter?

Read next Watch Rosetta's mesmerising journey around comet 67P Watch Rosetta's mesmerising journey around comet 67P

Yes, the rotation matters. Comet 67P has a rotation period of 12.7 hours. This means that if you were standing on the equator, you would be moving in a circle of radius 2πR in just 12.7 hours. This is the same as a linear velocity of 0.27 m/s.

So, at the equator, you don't even have to jump with a speed of 0.46 m/s to escape the comet. Ok, for the rest of the post I will assume an astronaut at the North pole where the comet rotation doesn't really matter.

What about skipping?

There you are on Comet 67P. You have been standing in the same spot and now you feel like you should move a little bit. You have to be super careful. If you move too fast, you will be gone. You must take just a super tiny hop.

Let's say you want to push off with a speed of just 0.01 m/s at a 20° angle. How far would you go horizontally? This is just a basic projectile motion problem (I hope). If I assume the vertical acceleration is constant then I can break this problem into both an x-motion (horizontal) and y-motion (vertical) problems. I will basically get the following kinematic equations (assuming I start at x = 0, y = 0).

Read next Farewell Philae: Lander tweets its heartbreaking goodbyes Farewell Philae: Lander tweets its heartbreaking goodbyes

I can find the initial x and y velocities with:

In order to find out how far the astronaut traveled (horizontally), I can first solve the vertical motion equation for the time since the final y position will also be zero (gets back to the ground).

Read next Gallery: The ice on comet 67P is as old as the solar system Gallery Gallery: The ice on comet 67P is as old as the solar system

Now I can use this expression for time in the x-motion equation to solve for the final x-position.

If I put in the values for the initial velocity, the angle and the gravitational field, I get a horizontal distance of 1.23 meters. Not too bad - but remember, this is at a launch speed of 1 cm per second. That's pretty slow. In fact, this one little "step" on a comet would take 269 seconds. That is long enough to sing a song or two - all in one leap.

What if you sneezed?

A skipping speed of 0.01 m/s is super slow. What if you are pushing off for your leap and you accidentally go just a little bit faster? What will happen?

Read next The ice on comet 67P is as old as the solar system The ice on comet 67P is as old as the solar system

Ok, no more playing around. No more assumptions of a constant gravitational field and a flat surface. Of course, this also means that I have a much more difficult projectile motion problem. Since the gravitational field can change in both magnitude and direction, I really need to use a numerical calculation where I use a computer to break the problem into many smaller steps.

If there is an interest, I can go over this program in another post - but for now let me just show you the results. This is a plot of the distance along the surface of the comet (the arc length) as a function of initial speed for the "skip" with a launch angle of 20°.

A launch speed of 0.15 m/s gets you a leap distance of about 300 meters. That's a serious leap. You probably didn't mean to go that far.

Could you jump into orbit?

Read next Our last chance to contact Philae lander probably failed Our last chance to contact Philae lander probably failed

Since it seems possible that an astronaut could jump faster than the escape velocity of the comet, could the astronaut instead jump into orbit around the comet? Maybe. I think the hardest problem would be getting in the right trajectory when starting from the surface.

Let me start with a circular orbit a distance of 10 meters above the surface. In a circular orbit, the gravitational force would have to be responsible for the acceleration of the astronaut moving in a circle. Since I know an expression for the acceleration of an object moving in a circle, I can write the following:

Here r is the distance from the centre of this spherical comet and the astronaut (so 2010 meters). I can now solve for the orbital velocity.

With the orbital altitude of 10 metres above the surface, I get an orbital speed of 0.32 m/s.

Now for the fun part. Let's launch our astronaut with a speed of 0.32 m/s from the surface of the comet and see what happens. I quickly wrote this program in Glowscript - you can take a look if you like.

Here is a view of the trajectory of the jumping astronaut.

You can see the astronaut does not make it into orbit.

What about friction?

I hate to say this, but my original plan was to see if there would be enough friction on the surface of this comet to be able to move.

Since the gravitational force is so small, there will be a very small frictional force between the astronaut's feet and the surface. The harder the astronaut pushes down on the surface, the greater this frictional force. But would the astronaut have to push so hard that the launch velocity would be too high? I'm not sure.


How about you look at the frictional forces on the skipping astronaut as a homework.

Back to the original question: could you walk on a comet? I don't think I would call it "walking". What about the 1998 movie Armageddon with Bruce Willis? In that movie, some dudes land on an asteroid and walk around while drilling holes. Without knowing the size and mass of the asteroid, I am going to guess that they couldn't walk on that object either. I know, it's a shock that Armageddon probably wasn't real.

This article originally appeared on Wired.com