- #1
jack476
- 328
- 125
Homework Statement
Two identical uniform metal spheres of radius 47 cm are in free space with their centers exactly 1 meter apart. Each has a mass of 5000 kg. Without integrating, show that gravity will cause them to collide in less than 425 seconds. [/B]
Source: Classical Mechanics, R. Douglas Gregory, chapter 4 "Problems in particle dynamics", problem 4.3.
Homework Equations
Standard formula for gravitational force and potential, definition of average force as impulse over time interval, and kinetic energy.
The Attempt at a Solution
I'm looking at how long it takes one of the spheres to travel the 3 cm so that their surfaces collide at the barycenter. The initial gravitational potential energy on one sphere due to the second is [tex] u_i = G\frac{(5000kg)^2}{1m} = 0.00167 J[/tex] and its final potential energy when the spheres have collided is [tex] u_f = G\frac{(5000 kg)^2}{0.94 m} = 0.00178 J[/tex] so the work done is [tex] W = 5.5*10^{-5} J [/tex] Since the potential energy was converted into kinetic energy, the final speed is given by [tex] V = \sqrt{\frac{2W}{M}} = 1.48*10^{-4} \frac{m}{s} [/tex] The impulse delivered to the sphere is therefore [tex] \Delta p = MV = 0.74 \frac{kgm}{s} [/tex] Suppose that [tex] \Delta t \geq 425 s [/tex] Then [tex] \langle F \rangle = \frac{\Delta p}{\Delta t} \leq 1.74*10^{-3} N[/tex]
With the formula for inverse square gravity, the initial force on the sphere was 1.67*10-3 N and the final force on the sphere was 1.89*10-3 N. Then the geometric mean of the initial and final forces is 1.777*10-3 N, so the average force is greater than this. Contradiction.
I'm just a little uncomfortable with that last part with the average force: that tells us about the average of the values of the force at the beginning of the end, but not necessarily throughout the whole process. But I can't think of any other directions to go in, because all of the other formulas that have been presented in the chapter that would be relevant to the collision time were obtained by integration, which I'm instructed not to use.