-1

Consider aspacecraft with mass $M=160\:\rm Mg$. An astronaut on the craft, in trying to stop the thruster, lost a tool with mass $m=250\:\rm g$. The tool now sits $2\:\rm m$ from (outer shell of) the spacecraft.

A question asks: If the tool is travelling with a speed of $2\:\rm m/s$ will gravity pull the tool back or is it lost forever?

Is it alright just to find the the escape velocity and compare it to the speed of the tool? The questions is worth a lot so I am confused.

gen-ℤ ready to perish
  • 6,610
mathsiscool
  • 29
  • This is a question more suited for physics.stackexchange.com. But yes, you're looking for the escape velocity. – Andrew Tindall Sep 03 '19 at 03:57
  • Is the spacecraft in orbit around a planet, or is it in an inertial frame? In what direction is the tool travelling? – Robert Israel Sep 03 '19 at 04:01
  • 1
    If the tool was lost while trying to stop the thruster then the tool didn't just start drifting away, the spaceship accelerated away from the tool until the thruster was finally stopped. Someone is not thinking through the problems they write. – David K Sep 03 '19 at 04:08
  • 1
    Setting aside the runaway thruster (let's suppose it stopped an instant before the tool was dropped), we don't know how deep in the spaceship's gravity well the outer skin is, therefore we can't know how deep the tool is or what its escape velocity is. My guess is the person who posed this question wanted you to make the absurd assumption that the distance from the outer shell can be treated as the distance from a point mass of $160\ \mathrm{Mg}$. – David K Sep 03 '19 at 04:12

1 Answers1

2

This should moved to Physics.SE, but I'm writing this to provide an answer for there after move completion.

Yes, what you want is the escape velocity from the spacecraft. However, finding that is actually quite tricky in this case: the spacecraft is an extended object! The usual formula for escape velocity,

$$v_\mathrm{esc} = \sqrt{\frac{2GM}{r}}$$

is based on the assumption that the object is either point-like or spherically symmetric. A real spacecraft is likely neither, and even in this case you can't necessarily assume a "spherical cow" because the radial distance $r$ in the case of an extended spherically-symmetric object must be measured from the center, not the surface, and you are given the distance from the spacecraft surface.

That said, the escape velocity for an extended mass distribution at a distance from its surface will, generally speaking, be lower than that same distance from it concentrated at a point, so you can use this as an upper bound on the escape speed.

Thus, given that in this system of units $G \approx 6.6743\ \times 10^{-8}\ \mathrm{\frac{kN \cdot m^2}{Mg^2}}$, then taking $M = 160\ \mathrm{Mg}$ and $r = 2\ \mathrm{m}$ gives the upper bound on $v_\mathrm{esc}$ as about 0.0033 m/s. Hence, yes, the screwdriver escapes easily.

(And this is all the more guaranteed by the fact that it is apparently thrusting away, which will cause it to leave in a hurry!)

The_Sympathizer
  • 19,651