On Earth, an astronaut and his space suit weigh 1960 N.  While working outside of the international space station weightless in space, the astronaut fires a 100 N rocket backpack for 3 seconds directed away from the station.  What is the resulting velocity of the astronaut and equipment and how far away from the iss is he when the rocket stops firing?  

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As I understand, the rocket backpack provides a constant force of `100 N.` During its work it throws away some mass, therefore the mass of the astronaut with his equipment changes. We don't know how it changes, but I suppose that this change is small (negligible).

The mass `m` of an astronaut with his space suit is his weight on Earth divided by Earth's gravitational acceleration `g=9.8 m/s^2.` His mass is `200 kg.` The mass remains the same in space, while weight may change even on Earth (during a free fall, for example).

By Newton's second law, the astronaut gets a constant acceleration of `a=F/m=100/200=0.5(m/s^2),` where `F` is a force. Therefore his velocity is `V(t)=at` and his displacement is `d(t)=(a t^2)/2` (because the initial speed is zero).

Thus, after `3` seconds the velocity will be `0.5*3=1.5 (m/s)` and the distance will be `(0.5*9)/2=2.25 (m).`


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