Armageddon

Classical Mechanics Level 3

If the Earth were suddenly stopped in its orbit, how long in days would it take for it to collide with the Sun?

Details and Assumptions

Regard the Sun and Earth as fixed point masses.
Mass of the Sun, $M=1.989\times 10^{30} \text{kg}$ .
Mass of Earth, $m=5.972\times 10^{24} \text{kg}$ .
Distance Earth to the Sun, $R=1.496\times 10^{8} \text{km}$ .
The gravitational constant, $G=6.674\times 10^{-11} \text{m}^3\text{kg}^{-1}\text{s}^{-2}$ .

The answer is 64.562.

1 solution

Rajat Raj
May 9, 2014

NO NEED OF VALUES

Instead of completely stopping the earth give a slight gentle push to EARTH It will follow elliptical path with a large semi major axis We know that T^2 is directly proportional to R^3 now let the time period be t and radius of earth's orbit be r Then if it follows new path the radius will be R/2 now, T^2/(365)^2=r^3/(r/2)^3 This will give T=365/2sqrt(2) This is the time period of the whole elliptical path now the time taken to collide with sun =T/2=64.71

here this should be noted that push should be gentle so that it would seem like the earth is falling towards the sun Also there no effect of earth's gravity on sun

Great solution! I believe precision (up to 5 decimals...) given in the problem's solution is way too much, and it should have asked to round up to greatest integer. One important physical reason for that are the relativity effects that are not totally negligible: velocity at a distance equivalent to the diameter of the sun is about 1/100 of speed of light. Relativity effects varry with (v/c)², which gives a deviation of magnitude 10e-4.

mat baluch - 6 years, 4 months ago

I needed a hideous integral equation. $T=\frac{\pi r_0^{1.5}}{2\sqrt{2G(M+m)}}$

So defenitely your solution is awesome.

Muhammad Arifur Rahman - 4 years, 10 months ago

Armageddon

The answer is 64.562.

1 solution

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