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An artist’s rendering of Philae landing on a comet.
An artist’s rendering of Philae landing on a comet. (credit: modification of work by “DLR German Aerospace Center”/Flickr)

Let’s define upward to be the + y -direction, perpendicular to the surface of the comet, and y = 0 to be at the surface of the comet. Here’s what we know:

  • The mass of Comet 67P: M c = 1.0 × 10 13 kg
  • The acceleration due to the comet’s gravity: a = ( 5.0 × 10 −3 m/s 2 ) j ^
  • Philae’s mass: M p = 96 kg
  • Initial touchdown speed: v 1 = ( 1.0 m/s ) j ^
  • Initial upward speed due to first bounce: v 2 = ( 0.38 m/s ) j ^
  • Landing impact time: Δ t = 1.3 s

Strategy

We’re asked for how much the comet’s speed changed, but we don’t know much about the comet, beyond its mass and the acceleration its gravity causes. However, we are told that the Philae lander collides with (lands on) the comet, and bounces off of it. A collision suggests momentum as a strategy for solving this problem.

If we define a system that consists of both Philae and Comet 67/P, then there is no net external force on this system, and thus the momentum of this system is conserved. (We’ll neglect the gravitational force of the sun.) Thus, if we calculate the change of momentum of the lander, we automatically have the change of momentum of the comet. Also, the comet’s change of velocity is directly related to its change of momentum as a result of the lander “colliding” with it.

Solution

Let p 1 be Philae’s momentum at the moment just before touchdown, and p 2 be its momentum just after the first bounce. Then its momentum just before landing was

p 1 = M p v 1 = ( 96 kg ) ( 1.0 m/s j ^ ) = ( 96 kg · m/s ) j ^

and just after was

p 2 = M p v 2 = ( 96 kg ) ( + 0.38 m/s j ^ ) = ( 36.5 kg · m/s ) j ^ .

Therefore, the lander’s change of momentum during the first bounce is

Δ p = p 2 p 1 = ( 36.5 kg · m/s ) j ^ ( −96.0 kg · m/s j ^ ) = ( 133 kg · m/s ) j ^

Notice how important it is to include the negative sign of the initial momentum.

Now for the comet. Since momentum of the system must be conserved, the comet’s momentum changed by exactly the negative of this:

Δ p c = Δ p = ( 133 kg · m/s ) j ^ .

Therefore, its change of velocity is

Δ v c = Δ p c M c = ( 133 kg · m/s ) j ^ 1.0 × 10 13 kg = ( 1.33 × 10 −11 m/s ) j ^ .

Significance

This is a very small change in velocity, about a thousandth of a billionth of a meter per second. Crucially, however, it is not zero.

Check Your Understanding The changes of momentum for Philae and for Comet 67/P were equal (in magnitude). Were the impulses experienced by Philae and the comet equal? How about the forces? How about the changes of kinetic energies?

The impulse is the change in momentum multiplied by the time required for the change to occur. By conservation of momentum, the changes in momentum of the probe and the comment are of the same magnitude, but in opposite directions, and the interaction time for each is also the same. Therefore, the impulse each receives is of the same magnitude, but in opposite directions. Because they act in opposite directions, the impulses are not the same. As for the impulse, the force on each body acts in opposite directions, so the forces on each are not equal. However, the change in kinetic energy differs for each, because the collision is not elastic.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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