How Long Would Earth Take to Collide with the Sun if It Stopped Rotating?

[amind]

Homework Statement

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

 

[SteamKing]

According to the OP, the Earth is 13 orders of magnitude more massive than the sun.

 

[amind]

According to the OP, the Earth is 13 orders of magnitude more massive than the sun.

Sorry for that ,fixed it now

 

[Chestermiller]

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

 

[amind]

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

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 anyone body , acceleration of the other is ,
a = f_1/m1 + f_2/m2

 

[SammyS]

Homework Statement

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

d²x/dt² is not the same as (dx/dt)² .

 

[Chestermiller]

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 anyone body , acceleration of the other is ,
a = f_1/m1 + f_2/m2

Ah. I see what you did. Thanks.

Chet

 

[amind]

d²x/dt² is not the same as (dx/dt)² .

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

 

[dauto]

Use Kepler's 3rd law comparing Earth's 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.

 

[Chestermiller]

Your starting equation is:

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

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

Chet

 

[dauto]

Your starting equation is:

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

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

Chet

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).

 

[Chestermiller]

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).

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

 

[dauto]

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

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

 

[Chestermiller]

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

$$x=x_0\cos^2θ$$

 

[dauto]

$$x=x_0\cos^2θ$$

You forgot to define Θ.

 

[Chestermiller]

You forgot to define Θ.

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

 

[D H]

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.

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.

 

[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.

 

[dauto]

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

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

 

[dauto]

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

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

 

[Chestermiller]

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.

Thanks a lot. Very cute trick.

Chet

 

[amind]

Thank you all for the help.
I found the solution using Kepler's third law as stated above.
But I am curious ,how can that integral be solved ?

 

[Chestermiller]

Thank you all for the help.
I found the solution using Kepler's third law as stated above.
But I am curious ,how can that integral be solved ?

Do the first integration to get the velocity (using either dx/dt as an integrating factor or using conservation of energy). Then use the trig substitution that I recommended, in which x₀ is equal to the initial distance.

Chet

 

[PeroK]

The differential equation didn't give me the same answer as Kepler's 3rd law. From Kepler I got:
$$t = \pi \sqrt{\frac{r_0^3}{GM}}$$
Starting with
$$r'' = -\frac{GM}{r^2}$$
I got:
$$(r')^2 = 2GM \frac{r_0 - r}{r_0 r}$$
Then, using the trig substitution, I got:
$$(cos^2(θ))θ' = \sqrt{\frac{GM}{2r_0^3}}$$
And, finally
$$t = \pi \sqrt{\frac{r_0^3}{8GM}}$$

Any ideas where I've gone wrong?

 

[dauto]

Nothing is wrong. The r₀ of the second method is the distance at aphelium while the r₀ of the first method is the semi major axis. One is twice as big as the other for that degenerate orbit. The cube of that factor of 2 explains your factor of 8.

 

[amind]

Thanks a lot , I got it now