# 2-body central force ( gravity ) problem

• jobendorfer
In summary, a comet is released from rest at a distance r_max from the sun, with no angular momentum. The sun can be assumed as a point mass. The time it takes for the comet to hit the sun can be calculated using the equation t = sqrt( m/2 ) * ∫ dr / sqrt( E - U(r) ) and assuming that the comet's initial kinetic energy is 0 and potential energy is negative due to the sun's mass. The limits of integration are r_max to 0 and the variable can be changed to u = r/r_max for a simpler function. Alternatively, Kepler's Third Law can also be used to calculate the time.
jobendorfer

## Homework Statement

A comet is released from rest at a distance r_max from the sun.
Therefore it has no angular momentum.
We can assume the sun is a point mass ( i.e. sun's radius is zero ).

How long does it take for the comet to hit the sun?

Let m = comet mass
let M_s = sun's mass
let G = gravitational constant
r = distance of comet from the sun, with origin at the sun.

## Homework Equations

t = sqrt( m/2 ) * ∫ dr / sqrt( E - U(r) )

## The Attempt at a Solution

Since the comet is released from rest, it initially has kinetic energy T = 0 and potential
U = G*M_s*m/r_max.

E = T + U = G*M_s*m/r_max.

U(r) = G*M_s*m/r

E-U(r) = (G*M_s*m)( 1/r_max - 1/r)

After grinding a bit I got:

t = 1/sqrt( 2*G*M_s) ∫ dr/sqrt( 1/r_max - 1/r )

with the limits of integration being r_max to 0.

This is where my eyes glazed over, since it's been 25 years since I've been near a calculus course. Any hints about how to proceed?

This is Taylor, _Classical Mechanics_, problem 8.21, part (c).

I am NOT in a class. I'm a software grunt trying to educate myself, any help offered would be greatly appreciated. This problem has really had me tearing my hair out for about 3 days.

Thanks,
John
(jobendorfer@cyberoptics.com)

Welcome to PF!

jobendorfer said:
Since the comet is released from rest, it initially has kinetic energy T = 0 and potential
U = G*M_s*m/r_max.

The potential energy of the comet is negative. U=-G*M_s*m/r_max.

jobendorfer said:
E = T + U = -G*M_s*m/r_max.

U(r) =-G*M_s*m/r

E-U(r) = -(G*M_s*m)( 1/r_max - 1/r)

You see, the kinetic energy is all right now, greater than zero.

jobendorfer said:
After grinding a bit I got:

t = 1/sqrt( 2*G*M_s) ∫ dr/sqrt(-1/r_max + 1/r )

with the limits of integration being r_max to 0.

Actually, v=-dr/dt as r decreases with t as the comet approaches the sun.There is an easy way if you have problems with an integral. Try to put it into wolframalpha.com. But before that, remove irrelevant constants and denote the necessary ones (r_max) by a simple letter. It is also advisable to change the variable so you get a simpler function. This case you can use u=r/r_max.

When you want to do the integration yourself, it is a long way... start with substituting sqrt(r_max/r-1)=u.ehild

Last edited:
Hildegard52 said:

?

ehild

Ehild,

Thanks for the welcome and for your help. Very much appreciated!

I've been lurking here off and on for years, I just tend to read and listen a lot more than I write -- for some strange reason. :-)

J.

Don't be shy! Come and talk.

How do you proceed with that integral?

ehild

I tried your suggestion u = r/r_max:

dr = r_max*du

Integral[ r_max*du/sqrt( 1/u*r_max - 1/r_max ) ]

With a bit of algebra:

r_max^1.5 * Integral [ du / sqrt( 1/u - 1 ) ]

I fed that integral into Mathematic 6's Integrate function as:

Integrate[ (1/u - 1)^-0.5, u]

What I got was:

(0.666667 (1.- 1. u)^0.5 u Hypergeometric2F1[1.5, 0.5, 2.5,
u])/(-1. + 1/u)^0.5

It's an answer, but ... wow. Not even sure what a "Hypergeometric" is.
So I'm thinking that there's got to be some clever transformation that gets rid of the 1/u.
Or I've bodged the algebra again. Take your (non-exclusive) pick!

But I did figure out where I went wrong with the minus signs on the gravity potential, so the day wasn't completely unproductive :-)

Regards,

John

Hi John

Sometimes you get the result in a very complicated form with these programs. With Wolframalpha, it ishttp://www3.wolframalpha.com/Calculate/MSP/MSP18861a0f7db9ai44e4be00000h0feidd94e0ec2a?MSPStoreType=image/gif&s=3&w=470&h=97
http://www.wolframalpha.com/input/?i=int(1/sqrt(1/x-1)+dx)

You can simplify and it results in something as pi/2 after inserting the boundaries 0 and 1.

You can use also 1/u as variable ... But try what I suggested sqrt(r_max/r-1)=u. Do not forget to transform dr too.

There is an other method to get the time: Think of Kepler's Third Law.ehild

Last edited by a moderator:

## 1. What is the 2-body central force problem?

The 2-body central force problem is a mathematical model used to describe the motion of two objects that are interacting with each other through a central force, such as gravity. It is also known as the two-body problem or the two-body central force (gravity) problem.

## 2. What are the assumptions made in the 2-body central force problem?

The 2-body central force problem assumes that the two bodies are point masses, that they are only influenced by the central force between them, and that the central force is directly proportional to the distance between the masses.

## 3. How is the 2-body central force problem solved?

The 2-body central force problem is solved using equations of motion, such as Newton's laws of motion and the law of universal gravitation. These equations are solved using mathematical methods, such as calculus, to determine the position, velocity, and acceleration of the two bodies at any given time.

## 4. What are the applications of the 2-body central force problem?

The 2-body central force problem has many applications in physics and astronomy. It is used to study the motion of planets around the sun, moons around planets, and satellites around Earth. It is also used to model the motion of binary stars and other celestial bodies.

## 5. Are there any limitations to the 2-body central force problem?

Yes, there are limitations to the 2-body central force problem. It assumes that the two bodies are point masses, which may not accurately represent real-world objects. It also does not take into account the effects of external forces or the gravitational pull of other nearby objects. Thus, it is an idealized model and may not always accurately predict the motion of two bodies in space.

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