# G-Force between 2 objects, find time?

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1. Jul 27, 2015

### Oxford365

1. The problem statement, all variables and given/known data
Spaceship mass=450000kg Initial Velocity= 300m/s y + 400m/s z
Asteroid mass = 2000kg Initial Velocity= -100m/s y - 50m/s z
Speed after 498m/s
Question: If the force between the objects is 5000N, how long will it take for spaceship to capture the asteroid? (Assume capture happens at constant force.)

2. Relevant equations
F=GMm/R^2
F=ma

3. The attempt at a solution
5000=(6.67x10^-11)(450000kg)(2000kg) / r^2
distance equals: .0035m
I know the accelerations are Spaceship: .0111 and Asteroid: 2.5
I'm lost after this. My guess is that the two objects come together with acceleration of 2.51 and must travel .0035m. .0035=1/2(2.51)t^2
t=.053s

2. Jul 27, 2015

### DEvens

Maybe the force is not gravitational? This is supported by the part where it says to assume the capture happens at constant force. I don't know what the force might be, but it does not matter. Maybe it's Star Trek tractor beams. Maybe it's a very long chain on a winch.

What would it mean for a space ship to "capture" an asteroid? It would mean something about the asteroid being held onto by the ship, yes?

What does it mean that the speed after is 498 m/s? It would be the speed of the ship plus asteroid. But what does that mean?

Consider that the ship starts at (300 m/s, 400 m/s). So what is the magnitude of velocity vector for the ship at the start?

This is basically a collision where the two objects stick together. The "stick together" is going to take a finite amount of time. That will tell you how much change of velocity each object undergoes. The force will then tell you how long it takes.

Collisions are usually easiest in the centre of momentum frame. In that frame the two masses start out with equal but opposite momentums. In this frame when the two objects collide and stick they will be at rest. You can very easily work out what the acceleration is in this frame, and how long it takes to bring the objects to rest.

3. Jul 27, 2015

### Oxford365

Yes it is a totally inelastic collision, and the 498 is the speed of both after they become one. Could you elaborate more on this centre of momentum frame?

4. Jul 27, 2015

### DEvens

5. Jul 28, 2015

### rcgldr

Both objects are initially moving away from Vc (velocity of the center of mass) in opposite directions, so you can simplify this to the speeds relative to Vc. Since the speeds are in opposite directions, then the initial "velocity" can be stated as dr/dt = Sa+Ss, where r is distance between objects, t is time, Ss is speed of space ship away from Vc and Sa is speed of asteroid away from Vc

The acceleration of objects back towards Vc: $\ddot{r}$ is a function of distance between objects. The magnitude of force on each object = GMsMa/r^2 (Ms is mass of space ship, Ma is mass of asteroid). So this will lead to calculating acceleration: $\ddot{r}$, velocity: $\dot{r}$, and time: t as a function of current distance: r, initial distance: r0, and initial velocity: v0.

Since no effective radius was stated for the objects, my guess is you are to consider them as point masses.

Looks like the original poster isn't following up on this. There are prior threads about the time it takes for two objects to collide due to gravity, but not with an initial velocity as in this problem.

Last edited: Jul 30, 2015