Gravitation Problem- Sun and Earth

In summary, The Earth would reach the sun in approximately 500 seconds if it was stopped in orbit and allowed to fall straight towards the sun. This can be calculated using Newton's law of gravitation and the center of mass of the system. Alternatively, Kepler's third law can also be used to find the time it takes for the Earth to reach the sun, taking into account the change in semi-major axis of the orbit.
  • #1
Quotexon
11
0

Homework Statement


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.

Homework Equations


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

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.
 
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  • #2
Quotexon said:

Homework Statement


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.

Homework Equations


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

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.
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
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
Quotexon said:
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.

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
Multiply both sides by dr/dt, and then integrate the resulting exact derivatives on both sides with respect to t.
 
  • #6
Dick said:
The hard way is to integrate the differential equation. The easy way is to think about Kepler's third law.

Hello Dick...

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

Dick said:
How has the semi-major axis of the orbit changed?

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
Tanya Sharma said:
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.

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.
 
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  • #8
Tanya Sharma said:
If the Earth and sun move in a straight line ,they are not in orbit anymore.
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.
 
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  • #9
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
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.
 

Related to Gravitation Problem- Sun and Earth

1. How does the Sun's gravity affect the Earth's orbit?

The Sun's gravity is the dominant force that keeps the Earth in its orbit. The Sun's enormous mass creates a gravitational pull that keeps the Earth in a stable elliptical orbit around it. Without the Sun's gravity, the Earth would fly off into space in a straight line.

2. What is the force of gravity between the Sun and Earth?

The force of gravity between the Sun and Earth is approximately 3.52 x 10^22 Newtons. This is calculated using Newton's Law of Universal Gravitation, which states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them.

3. How does the distance between the Sun and Earth affect the strength of gravity?

The strength of gravity between the Sun and Earth is inversely proportional to the square of the distance between them. This means that as the distance between the two objects increases, the force of gravity decreases. This is why the Earth experiences different levels of solar gravity throughout its orbit around the Sun.

4. What is the effect of the Earth's mass on its orbit around the Sun?

The Earth's mass has a minimal effect on its orbit around the Sun. This is because the Sun is significantly more massive than the Earth, so its gravitational pull is much stronger. However, the Earth's mass does play a role in determining the speed of its orbit around the Sun. The more massive an object, the more inertia it has, which affects its speed and orbit.

5. How does the force of gravity between the Sun and Earth change over time?

The force of gravity between the Sun and Earth remains relatively constant over time. It is only affected by changes in the mass or distance of either object. However, the Earth's orbit around the Sun does change slightly over time due to factors such as the gravitational pull of other planets and the Sun's own movement within the galaxy.

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