Timeline for How does gradient give $g$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2020 at 19:15 | answer | added | R.W. Bird | timeline score: 0 | |
| Dec 14, 2020 at 16:19 | answer | added | The_Sympathizer | timeline score: 2 | |
| Dec 14, 2020 at 15:46 | vote | accept | protectgoodlivingbeingask | ||
| Jan 15, 2022 at 23:58 | |||||
| Dec 14, 2020 at 15:24 | comment | added | protectgoodlivingbeingask | Yes I knew that. | |
| Dec 14, 2020 at 15:24 | comment | added | protectgoodlivingbeingask | Ok now I understand.Yes I know I should have studied it a bit more before asking but . | |
| Dec 14, 2020 at 15:22 | comment | added | Vulgar Mechanick | are you aware how potential energy is defined for a general conservative force? | |
| Dec 14, 2020 at 14:52 | comment | added | Nihar Karve | I believe OP is asking this having seen that given a spherically potential: $V(r) = -\frac{GM}{r}$, the gravitational field strength is $\frac{GM}{r^2}$ - OP, please note that this is just a special case of $\nabla V$ = $\frac{\partial}{\partial r}V(r) =\frac{GM}{r^2}$, the former formula does not hold for general $V$ | |
| Dec 14, 2020 at 14:04 | comment | added | protectgoodlivingbeingask | As I am new to calculus could you please elaborate why I cannot just divide v by r=$\sqrt{x^2+y^2+z^2}$ and instead $\nabla V to g$. | |
| Dec 14, 2020 at 14:04 | answer | added | Karol | timeline score: 2 | |
| Dec 14, 2020 at 14:01 | comment | added | Charlie | The gravitational force is a conservative vector field so it can be written as the gradient of a potential function. | |
| Dec 14, 2020 at 13:51 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 2 characters in body; edited tags; edited title; edited tags; edited tags
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| Dec 14, 2020 at 13:44 | comment | added | protectgoodlivingbeingask | Here V varies with distance from (0,0,0). | |
| Dec 14, 2020 at 13:43 | history | asked | protectgoodlivingbeingask | CC BY-SA 4.0 |