I know that to find the gravitational force between two objects, if either of them is a sphere, we can assume its mass to be concentrated at its center and use the formula for gravitational forces for point masses. However, in my textbook (Cengage Physics, Mechanics - II), to find the gravitational force exerted by the Earth on masses at the equator and poles (where they showed the Earth as a spheroid), they used $\frac{GM_E m}{R_{eq}^2}$ and $\frac{GM_E m}{R_{pole}^2}$ respectively. But I think that is wrong because the Earth, being treated as a spheroid in this case, cannot exhibit the behavior that spherically symmetric masses do.
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$\begingroup$ Related: physics.stackexchange.com/q/8074/2451 and links therein. $\endgroup$– Qmechanic ♦Commented May 18 at 9:33
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You are correct. Gravity around a spheroid is not the same as gravity around a sphere. The J2 perturbation accounts for the Earth's ellipsoidal shape, and technically one has to account for it. However, it is reasonably small so can often be ignored. Satellite operators, however, do need to account for this effect. Indeed, when you really look at it, the Earth's gravitational field is anything but simple
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$\begingroup$ Thank you for answer. Can you please help me with this problem too. $\endgroup$ Commented May 18 at 13:00
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$\begingroup$ It's easy to visualize centripetal force as a force that causes an object to be in circular motion in cases like rotating a stone by tying a string to it and an electron rotating around a nucleus because of electrostatic attraction acting between them as the centripetal force. But it is hard to visualize centripetal force in cases where a biker is moving his bicycle in a circular path because I think in this case rotational motion is simply exhibited due to the changing of the position of the handle and there's no centripetal force involved. $\endgroup$ Commented May 18 at 13:01
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$\begingroup$ The centripetal force in the biker's case comes from friction, but it's a bit harder to see. Bike tires are rounded, so leaning the bike causes the inside of the contact patch to move less per revolution than the outside of the contact patch. The only no-slip solution for that contact patch is a circular path. If the bike "tries" to go straight, it stretches the rubber, generating a inward force that puts the bike on that circular path $\endgroup$ Commented May 18 at 15:43
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$\begingroup$ The handlebars are a bit more complicated. At low speeds (15mph or less, give or and take), turning the handlebars just turns the wheel to change the direction the front wheel wants to go. Faster than that, the centrifugal force would throw you off to the outside (a kind of fall called "high siding" which is very dangerous). So you have to lean inward. Now turning in the direction of the turn causes you to lean less, and turning outward causes you to lean more, paradoxically making you turn sharper. We call it counter steering, and all bikers do it at speed. $\endgroup$ Commented May 18 at 15:49