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

• amind
In summary, the conversation discusses how long it would take for the Earth to collide with the Sun if the Earth suddenly stops rotating around the Sun. Various equations and methods are proposed, including using Kepler's third law, energy conservation, and trigonometric substitutions, to solve the problem. The discussion also touches on the relationship between gravitational forces, mass, and acceleration.
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

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

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

Sorry for that ,fixed it now

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

Chestermiller said:
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

1 person
amind said:

## 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 .

amind said:
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

SammyS said:
d2x/dt2 is not the same as (dx/dt)2 .

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

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.

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

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

Chet

Chestermiller said:

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

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).

Last edited:
dauto said:
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

Chestermiller said:
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

dauto said:
It can be done fairly easily if you know a couple of tricks of the trade
$$x=x_0\cos^2θ$$

Chestermiller said:
$$x=x_0\cos^2θ$$

You forgot to define Θ.

dauto said:
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

Chestermiller said:
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.

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

Chestermiller said:
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.

Pranav-Arora said:
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.

D H said:
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

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 ?

amind said:
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

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?

1 person
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 and PeroK
Thanks a lot , I got it now

## 1. What is the "Earth falling into sun problem"?

The "Earth falling into sun problem" refers to the hypothetical scenario in which Earth's orbit around the sun decays to the point where it eventually collides with the sun, leading to the destruction of our planet.

## 2. How likely is it that Earth will fall into the sun?

Based on our current understanding of planetary motion and the laws of physics, it is highly unlikely that Earth will ever fall into the sun. The sun's immense gravitational pull keeps Earth in a stable orbit, and it would take an extraordinary event or force to disrupt this balance.

## 3. What factors could potentially cause Earth to fall into the sun?

The most significant factor that could potentially cause Earth to fall into the sun is a significant decrease in its orbital velocity. This could be caused by a collision with a large object, a close encounter with another planet, or a major disruption in the solar system's overall gravitational forces.

## 4. Is there anything we can do to prevent Earth from falling into the sun?

As mentioned earlier, the chances of Earth falling into the sun are incredibly slim. However, if such a scenario were to become a possibility, there are theories and technologies that could potentially be used to alter Earth's orbit and prevent a collision. These include gravitational assists from other planets, solar sails, and nuclear propulsion systems.

## 5. How long do we have until Earth would fall into the sun?

It is impossible to predict an exact timeline for when Earth may potentially fall into the sun. However, it is estimated that the sun will become a red giant in about 5 billion years, which would likely engulf and destroy Earth. But even before that, it is highly unlikely that Earth's orbit will decay enough to cause a collision with the sun within the next several million years.

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