Without integrating, show two metal spheres collide in <425s

Click For Summary

Homework Help Overview

The problem involves two identical metal spheres in free space, initially 1 meter apart, and requires demonstrating that gravity will cause them to collide in less than 425 seconds without using integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the time until collision by analyzing gravitational potential energy and the resulting kinetic energy. They express uncertainty regarding the average force calculation over the entire process.
  • Some participants suggest simplifying the problem by considering the distance each sphere must travel and assuming a constant gravitational force for the calculation of time until collision.
  • Questions arise about the implications of assuming constant force and how it affects the accuracy of the collision time estimate.

Discussion Status

Participants are exploring different methods to approach the problem, with some providing calculations based on assumptions of constant force. There is acknowledgment that the actual collision time would be less than the calculated estimate due to increasing acceleration.

Contextual Notes

The original poster is instructed not to use integration, which influences the methods and reasoning discussed in the thread.

jack476
Messages
327
Reaction score
124

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 u_i = G\frac{(5000kg)^2}{1m} = 0.00167 J and its final potential energy when the spheres have collided is u_f = G\frac{(5000 kg)^2}{0.94 m} = 0.00178 J so the work done is W = 5.5*10^{-5} J Since the potential energy was converted into kinetic energy, the final speed is given by V = \sqrt{\frac{2W}{M}} = 1.48*10^{-4} \frac{m}{s} The impulse delivered to the sphere is therefore \Delta p = MV = 0.74 \frac{kgm}{s} Suppose that \Delta t \geq 425 s Then \langle F \rangle = \frac{\Delta p}{\Delta t} \leq 1.74*10^{-3} N

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.
 
Physics news on Phys.org
It's simpler than you think.
1. How far does each sphere have to travel before the collision?
2. If the force between spheres is assumed to be constant at its initial value, how long will it take for the spheres to collide?
 
Thank you for your reply and sorry for the late reply.
kuruman said:
It's simpler than you think.
1. How far does each sphere have to travel before the collision?

The center of each sphere travels 0.03 meters.
2. If the force between spheres is assumed to be constant at its initial value, how long will it take for the spheres to collide?

If the force is constant at its initial value of 1.67*10-3 N then the constant acceleration is a = 3.34*10-7 m/s2. Then the equation of its path is x(t) = x0 - (a/2)t2, and x0 = 0.03. The positive root of this equation is t = 423.84 seconds, so that would be the travel time. Since the acceleration would actually have been increasing, the real travel time would be less than this. So t < 425 seconds.

Thanks for the help!
 
Well done!
 

Similar threads

  • · Replies 97 ·
4
Replies
97
Views
6K
Replies
2
Views
3K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
15
Views
2K
Replies
17
Views
2K
Replies
7
Views
1K
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K