came from your hohmann transfer looking through a par-e light problem I was having.... I was throwing be-borlie, hisinberg, boltsman and showdinnger lol. but this and your vid on hohman transfer really helped
in your video, there 2 potential energy formulas: U=mgh and U=-GMm/r. you should show your reader that, the first one is based on setting the surface of the earth as zero point, (of course, it is approximate as it uses g = GMm/R2 to replace GMm/(R+h)xR while the second one set zero at infinity far. This is the key point of the calculation.
You bring up a good point. I left out the zero point for U=mgh because if using that fxn, you can choose any point to be zero, not just the ground level. However I certainly could have got to the point faster at the end when discussing the zero point for U=-Gmm/r as well as leading into escape velocity.
Essentially. The only use of the integral is to show the relationship between Fg and Ug. But BE CAREFUL, unlike with mgh, you cant just plug in a change in height for h, you have to evaluate Ug at two different r's.
came from your hohmann transfer looking through a par-e light problem I was having.... I was throwing be-borlie, hisinberg, boltsman and showdinnger lol. but this and your vid on hohman transfer really helped
in your video, there 2 potential energy formulas: U=mgh and U=-GMm/r. you should show your reader that, the first one is based on setting the surface of the earth as zero point, (of course, it is approximate as it uses g = GMm/R2 to replace GMm/(R+h)xR while the second one set zero at infinity far. This is the key point of the calculation.
You bring up a good point. I left out the zero point for U=mgh because if using that fxn, you can choose any point to be zero, not just the ground level. However I certainly could have got to the point faster at the end when discussing the zero point for U=-Gmm/r as well as leading into escape velocity.
great video👌🏿
Thanks!
so basically it’s Fg.h but h is r and Fg is the Newton’s law form Fg = -GMm/r2 and so Ug = -GMm/r right ? no need to use integral or stuff
Essentially. The only use of the integral is to show the relationship between Fg and Ug.
But BE CAREFUL, unlike with mgh, you cant just plug in a change in height for h, you have to evaluate Ug at two different r's.
so basically it is Fg.h after all
Only if you are lifting something an infinitely small distance.