# Gravitation Problem- Sun and Earth

1. Dec 2, 2012

### Quotexon

1. The problem statement, all variables and given/known data
The earth orbits around the sun because it has angular momentum. If we
stopped the earth in orbit and then let it fall straight towards the
sun, how long would it take to reach the sun in seconds?

Details and assumptions

The mass of the sun is 2×10^30 kg.
The mass of the earth is 6×10^24 kg.
Newton's constant is 6.67×10^-11 Nm^2/kg^2.
The earth is 149,600,000 km from the sun.
You may treat the earth and sun as point masses.

2. Relevant equations
F = GMsMe/R^2
center of mass R= (MsRs + MeRe)/(Ms + Me)
d^2r/dt^2 = GMs/r^2

3. The attempt at a solution
I realized that as these point masses get closer, the distance between them decreases resulting in a stronger force and thus a greater acceleration.
d^2r/dt^2 = GMs/r^2
Since the Sun is also mutually attracted by the Earth and may move slightly, would using the center of mass help in any way?
I'm not sure if this is the right way to approach this. Any help would be greatly appreciated.

2. Dec 2, 2012

### SammyS

Staff Emeritus
Yes, using the center of mass of the system will help.

The earth will collide with the sun when the earth reaches the center of mass.

Remember, that the r in the Law of universal gravitation is the distance between the two bodies, not earth's distance from the center of mass.

3. Dec 2, 2012

### Quotexon

How would I resolve the fact that the acceleration changes over time as a function of the distance decreasing? I'm not too solid on the possible mathematics that that might entail. Thanks very much.

4. Dec 2, 2012

### Dick

The hard way is to integrate the differential equation. The easy way is to think about Kepler's third law. How has the semi-major axis of the orbit changed?

5. Dec 2, 2012

### Staff: Mentor

Multiply both sides by dr/dt, and then integrate the resulting exact derivatives on both sides with respect to t.

6. Mar 27, 2014

### Tanya Sharma

Hello Dick...

Could you please explain how Kepler's third law is applicable in this case .

I think if the earth is stopped in orbit ,it would fall towards the sun .The sun would also move towards the earth .Both will collide at their COM .

If the earth and sun move in a straight line ,they are not in orbit anymore.Then how to answer your question .

I would be grateful if you could explain this concept.

7. Mar 28, 2014

### Dick

Sure, but before they collide, you make the pretty good approximation that the sun is infinitely massive and the earth orbits around it. Or if you want to be a little less sloppy you use the notion of reduced mass, http://en.wikipedia.org/wiki/Reduced_mass, to change two body motion into the motion of one body around a fixed potential. Treat the straight line as a degenerate ellipse.

8. Mar 28, 2014

### Staff: Mentor

A straight line collision or escape trajectory is still a valid conic section, so the relevant math still applies. These linear cases are sometimes referred to as degenerate orbits.

9. Mar 28, 2014

### BruceW

yeah, the Kepler's Law method is a much nicer way to find the answer (assuming the mass of the sun is much bigger than the mass of the earth, which it is). and I just checked that it does give the same answer as doing the integration - mostly to reassure myself :) So, if the OP'er is not comfortable using equations with non-constant acceleration, then it's probably best to use the Kepler's third law method.

10. Mar 28, 2014

### dauto

There is another live thread on exactly that very same topic where both methods have already been discussed in more detail. I'm not sure about how to link there.