Time taken for a planet to collide with the Sun

In summary, the time taken for a planet to collide with the Sun varies based on its initial distance, orbital velocity, and gravitational influences. Generally, a planet's trajectory can be altered by factors such as gravitational interactions with other celestial bodies, leading to different potential collision timelines. Calculations and simulations show that, under certain conditions, a planet could take millions to billions of years to spiral into the Sun, depending on its specific orbital characteristics.
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someone_2156
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TL;DR Summary: if a planet was suddenly stopped in its orbit, suppose to be circular, find how much time will it take in falling onto the sun(in terms of time period)

I know that T=2pi (r^3/GM)^1/2 and a=GM/x^2
how to proceed further
 
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Your approach won't work because the ##x=c_1+c_2t+\frac 12at^2## approach you are trying only works if ##a## is constant. You would need to solve the differential equation you started with. Deciding whether you are using ##x## or ##R## for the planet's position will help with that, but it produces a fairly nasty answer.

Are you familiar with Kepler's laws?
 
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What would happen to the period of the orbit if you abruptly slowed, rather than stopped, the planet? And if you did it again but slowed it more? What is the limit of that if you stop the planet?
 
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Also, you may want to read the LaTeX Guide linked immediately below the reply box if you are posting maths.
 
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Ibix said:
What would happen to the period of the orbit if you abruptly slowed, rather than stopped, the planet? And if you did it again but slowed it more? What is the limit of that if you stop the planet?
i am not sure about this. i know that t^2 is directly proportionl to the cube of semi major axis
 
  • #6
someone_2156 said:
i am not sure about this. i know that t^2 is directly proportionl to the cube of semi major axis
There are two ways to solve the problem. The first involves solving a differential equation - which is quite hard. The second is a neat trick using Kepler's laws. It's not easy, however, to figure out the trick if you haven't seen it before.

Where did you get this problem and what level of physics are you studying?
 
  • #7
PeroK said:
There are two ways to solve the problem. The first involves solving a differential equation - which is quite hard. The second is a neat trick using Kepler's laws. It's not easy, however, to figure out the trick if you haven't seen it before.

Where did you get this problem and what level of physics are you studying?
undergraduate/ secondary school and I got this problem from a question bank. Can you explain the second method?
 
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someone_2156 said:
undergraduate/ secondary school and I got this problem from a question bank. Can you explain the second method?
What happens if, instead of stopping the planet, you leave the planet with a small tangential velocity? Imagine the star and planet are point masses.
 
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  • #9
PeroK said:
What happens if, instead of stopping the planet, you leave the planet with a small tangential velocity? Imagine the star and planet are point masses.
the orbit would flatten more and the closest approach to the sun would be very near
 
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someone_2156 said:
the orbit would flatten more and the closest approach to the sun would be very near
Try to do some calculations using Kepler's laws.
 
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someone_2156 said:
the orbit would flatten more and the closest approach to the sun would be very near
The key thing to think about is where the foci of the ellipse move to. In the circular case, both foci are at the center. Where are they in the extremal case as the circle flattens more and more?

Hint: there's probably a tutorial on the web on how to draw an ellipse with string and a pair of pins.
 
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