Calculations of Earth falling into the Sun if it suddenly stopped orbiting?

Question

If the earth stopped orbiting the sun, it would be pulled by the Sun's gravity and fall into it. How much time it will take is tricky as force is inversely proportional to square of distance. So I have solved this problem using calculus. See my solution in this picture. My solution gives the right answer, but have I set up it the right way? And are my assumptions right?

your equation dr=1/2a t^2 is wrong, both sides have different orders. You should have used dr=vdt, but this would not be useful anyways. You need to integrate newton's second law, which is different than the ones you used. Are you sure you got the right answer? And you cannot integrate $dt^2$ — , Dec 25 '20 at 18:01
My solution gives the right answer. No it doesn’t. Your numerical prefactor is incorrect, although it happens to be close to the correct number. — G. Smith, Dec 25 '20 at 18:14
Incidentally, there is a much easier way to solve this problem, without any direct integration, using only Kepler's Third Law. — Buzz, Dec 25 '20 at 18:14
Please be aware that check-my-work questions are off-topic here. — G. Smith, Dec 25 '20 at 18:16
Does this answer your question? What if the Earth stops in its orbit and instead falls straight towards the Sun? — , Dec 25 '20 at 18:35
Please do not post images of text and math. They are not searchable and not accessible to visually-impaired members. You are expected to type all text and use MathJax for all math. — G. Smith, Dec 25 '20 at 18:39

score -2 · Answer 1 · answered Dec 25 '20 at 18:06

-2

your statement $$ dr=\frac{a}{2}dt^2 \text{ is not good , you could derive } \frac{dr}{dt}=\frac{a}{2}dt$$

answered Dec 25 '20 at 18:06

trula

6,146

score -3 · Answer 2 · answered Dec 25 '20 at 18:30

$\frac{dr}{dt}$ is not a fraction. So, you need to be careful how to handle it. This equation , $dr = \frac{1}{2} a dt^2$ is wrong because at one side you have $dr$ and the other side you have $dt^2$.

Here is how you should proceed.

$$\ddot r = \frac{GM}{r^2}$$

$$\frac{d^2r}{dt^2} = \frac{GM}{r^2}$$

Note that this is a second order non-linear differential equation. You can see how to solve this differential equation here https://math.stackexchange.com/questions/681838/what-went-wrong-one-dimensional-inverse-square-law

It’s much easier to start with the conservation of energy. — G. Smith, Dec 25 '20 at 18:33

Calculations of Earth falling into the Sun if it suddenly stopped orbiting?

2 Answers2