Question: ( b ) Say the force applied by the space - ship is just gravitational attraction. If the asteroid has mass ( M _

(b) Say the force applied by the space-ship is just gravitational attraction. If the asteroid has mass \( M_{a}\) and the space-ship has mass \( M_{s}\) and they are separated by a distance \( d \) then the gravitational force between them is \( G M_{a} M_{s}/ d^{2}\) where \( G=6.67\times 10^{-11}\mathrm{Nm}^{2}/\mathrm{kg}^{2}\) is the gravitational constant. (Note to actually keep this force constant on the asteroid, the space-ship will have to keep the distance \( d \) constant against the pull of the asteroid's gravity, so it will take some jet propulsion to hold that distance fixed.) To keep the ship safe from any other debris traveling with the asteroid (e.g., knocked off by small collisions), it's best to keep \( d=10\mathrm{~km}\).
i. What is the mass of the ship needed to have the asteroid miss the Earth by both distances in the first part of the problem?
ii. If most of this mass is just rock loaded onto the ship from the Moon (from where it's easier to launch) then what would be the size of the ship in each case, assuming it's basically a cube (i.e. what is the length, width or height of the ship)? Assume lunar rock density is \(3000\mathrm{~kg}/\mathrm{m}^{3}\). Are these plausible or implausible ship sizes (e.g., is the ship the size of a planet - which makes it implausible - or the size of a building or an air-craft carrier - which is more plausible)?
( b ) Say the force applied by the space - ship

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Physics Questions!