Force of Gravity and Gravitational Potential Energy Functions from Zero to Infinity (APC Mechanics)

[@RLakshman1916]

Wow, excellent video! I love your channel, makes me love physics more! The effort you put in your videos is incredible. Keep going!

[@FlippingPhysics]

Thanks! It warms my heart to know I help you love physics.

[@mr.swaney8300]

Awesome video, as always! I was thinking... since we're assuming the planet is spherical and that it has a constant density, the density of the planet is Mp/Vp = Mp / (4/3 pi Rp^3). If you plug that in you can find that Ug at r=0 is -3/2 * G Mp m0 / Rp. How awesome is that! Therefore, if you were to take the object from the very center of the planet to very far away from the planet you'd do exactly 1/3 of your work getting it to the surface and 2/3 getting it the rest of the way to infinity. I think plugging in the density of the planet the whole way through makes so much cancel out that it actually makes the derivation easier to look at.

[@swiftyedits4067]

Good video. Perhaps you should label the titles so that students know this is for AP Physics C-physics 1 students would be confused

[@FlippingPhysics]

Valid point. The description does start with the word _calculus_ which should give a good hint. However, it does not hurt to put in the title as well. I have now done so. Thanks.

[@_BhagavadGita]

Excellent, as always, Mr P.

[@FlippingPhysics]

Thank you.

[@BrunovonSilberberch]

Can you make some videos on the recently released 2017 AP Physics 1 response questions.

[@FlippingPhysics]

Not currently planning on doing so. I am concentrating on finishing AP Physics 1 topical videos first. Just so you know, Dan Fullerton has them done already. kzbin.info/www/bejne/aqDKZqmjedR3pJI

[@raghu7463]

Sir could you please give me the solution of this question The acceleration due to gravity at the latitude 45 degrees on the earth becomes zero if the angular velocity of rotation of earth is.

[@carultch]

At 45 degrees latitude, no amount of rotational speed will make it possible for the apparent gravitational field to equal zero. The true gravity acts radially inward toward the center of Earth, and the centripetal acceleration is radially inward toward the axis of Earth, perpendicular to the axis. The only location where these vectors align is at the equator. You can equate G*M/r^2 * cos(phi) = omega^2*r, and solve for omega where the centripetal acceleration completely "uses up" as much gravity as it possibly can. However, you are still left with G*M/r^2 * sin(phi) acting parallel to the axis and toward the equator. Since there is no dependence on omega in this term, you cannot nullify it by changing the rotational speed. If this happened on a spherical Earth, assuming it remains spherical, you would feel like you are on a 45 degree hill, when standing on what once was level ground, and your apparent weight would be 71% of your weight today. The only location where thought experiment works, is at the equator. You end up with a period of rotation of 88 minutes, which is the minimum possible orbital period of satellites.

[@_BhagavadGita]

Mr P: Is it correct to say that if the planet is of uniform density and spherically symmetric, then these equations would still be valid even if the planet were rotating about the axis defined by the tunnel?

[@FlippingPhysics]

If the axis is defined by the tunnel, I do not see how a rotating planet would affect this question.

[@tapwater311]

i luv it

[@FlippingPhysics]

Glad I am.