# Earth falling into sun problem

1. Mar 26, 2014

### amind

1. The problem statement, all variables and given/known data
If Earth stops rotating Sun suddenly , how long would it take for Earth to collide with Sun ?
G = gravitational constant
m_s = 2e30kg
m_e = 6e24kg
where m_s is the mass of the sun and m_e , mass of earth
x = 1au =1.5e11m
2. The attempt at a solution
Gravitational acceleration ,
a = G(m_s + m_e)/x^2

d^x/dt^2 = a
dt = sqrt(1/a).dx
dt = x/sqrt(G(m_s + m_e)) . dx
integrating over x from x = distance between earth and sun to 0
t = x^2/(2*sqrt(G(m_s + m_e)))

The answer i m getting is not right , anyone help me finding out the solution and what i did wrong

Last edited: Mar 26, 2014
2. Mar 26, 2014

### SteamKing

Staff Emeritus
According to the OP, the earth is 13 orders of magnitude more massive than the sun.

3. Mar 26, 2014

### amind

Sorry for that ,fixed it now

4. Mar 26, 2014

### Staff: Mentor

Your equation for the attractive force is incorrect. It should involve the product of the masses, not the sum. The acceleration should be independent of the mass of the earth.

Chet

5. Mar 27, 2014

### amind

Gravitational forces act on both objects , both are directed towards each other (they finally collide at their centre of mass).
If we switch our reference frame to an inertial frame of any one body , acceleration of the other is ,
a = f_1/m1 + f_2/m2

6. Mar 27, 2014

### SammyS

Staff Emeritus
d2x/dt2 is not the same as (dx/dt)2 .

7. Mar 27, 2014

### Staff: Mentor

Ah. I see what you did. Thanks.

Chet

8. Mar 27, 2014

### amind

OK thanks for that , but how should I solve this problem then.

9. Mar 27, 2014

### dauto

Use Kepler's 3rd law comparing earths actual orbit (semi-major axis = 1 AU) with the earth's orbit in the problem (semi-major axis = .5 AU), and keep in mind that the time to fall into the sun is just half an orbital time.

10. Mar 27, 2014

### Staff: Mentor

$$\frac{d^2x}{dt^2}=-G\frac{(m_s+m_e)}{x^2}$$

If this is correct, then dx/dt is an integrating factor for both sides of your equation.

Chet

11. Mar 27, 2014

### dauto

The second integration is a bit more challenging though. It is easier to use Kepler's 3rd law and bypass the need to integrate at all (Well the integrations are hidden under the hood of the law).

Last edited: Mar 27, 2014
12. Mar 27, 2014

### Staff: Mentor

I hadn't looked at the second integration yet. I'll take a look at it later when I have some time. Is it just a bit more challenging, or is really much more challenging?

Chet

13. Mar 27, 2014

### dauto

It can be done fairly easily if you know a couple of tricks of the trade

14. Mar 27, 2014

### Staff: Mentor

$$x=x_0\cos^2θ$$

15. Mar 27, 2014

### dauto

You forgot to define Θ.

16. Mar 27, 2014

### Staff: Mentor

It was just a trigonometric substitution to facilitate the integration. But, I'd be interested in your filling me in regarding its relationship to Kepler's 3rd law.

Chet

17. Mar 27, 2014

### D H

Staff Emeritus
It's fairly simple, conceptually. Imagine a sequence of orbits all with aphelion distance equal to one astronomical unit but with perihelion distances tending to zero. In the limit, the orbiting object is falls straight toward the Sun from aphelion (r=1 AU), makes a 180 degree turn at perihelion (r=0), and climbs back to aphelion. By symmetry, the time taken in going from aphelion to perihelion and the time taken in going from perihelion to aphelion are both half an orbit.

That said, I suspect that the OP is supposed to integrate the equations of motion for this radial trajectory.

18. Mar 28, 2014

### Saitama

Energy conservation is also a way to solve the problem. You have to solve a first order differential equation if energy conservation is used.

19. Mar 28, 2014

### dauto

Oh, OK, Yes that substitution works. DH already explained the relationship with Kepler's law.

20. Mar 28, 2014

### dauto

That bypasses the 1st integration leaving the second integration to be done by hand.