Timeline for Deriving energy for elliptical orbit
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2022 at 23:40 | vote | accept | Anandatheertha Bapu | ||
| Dec 2, 2022 at 6:39 | comment | added | Mark H | @AnandatheerthaBapu At the closest distance, $GMm/r^2 < mv^2/r,$ so the planet starts moving away from the star. At the farthest distance, $GMm/r^2 > mv^2/r,$ so the planet starts falling back towards the star. | |
| Dec 2, 2022 at 6:36 | comment | added | Mark H | @AnandatheerthaBapu Since the planet is at its closest distance from the star, it is moving perpendicular to the line joining the planet and star. So, there is zero tangential force; gravity pulls purely radially. The only force is gravity, so that radial component must be $GMm/r^2.$ Because this force is less than $mv^2/r,$ the planet will move in a wider arc than a circle with a radius of $r.$ That's why it will start to move farther from the star. | |
| Dec 2, 2022 at 4:00 | comment | added | Anandatheertha Bapu | Thank you very much. But what I still don't fully understand is what exactly the radial component of the force would be in this equation? I always thought that the radial component was mv^2/r and the tangential component of the acceleration was m$\alpha$r. | |
| Dec 2, 2022 at 3:41 | history | answered | Mark H | CC BY-SA 4.0 |