A satellite follows the elliptical orbit shown in FIGURE P12.77. The only force on the satellite is the gravitational attraction of the planet. The satellite's speed at point 1 is 8000 m/s. a. Does the satellite experience any torque about the center of the planet? Explain.

Video transcript

Hey, everyone in this problem, we're told that mercury has the most elliptical orbit of any planet in the solar system. Mercury orbits the sun which is depicted in the diagram. We have the sun and mercury are attracted to each other by gravitational force. I want to ask, is there any torque encountered by Mercury due to the gravitational force around the center of the sun? And we need to justify our response. So if we look at our diagram here, we have this elliptic orbit with mercury going around, we have the sun and let's recall what the torque is. OK. We're given four answer choices here A through D and we're gonna go through them one at a time. But let's first write down what that torque is. OK. And recall that the torque tow can be written as R multiplied by F multiplied by sine. The A and R is the kind of radius vector pointing from son to mercury. OK. F is the force and then we have sign data. Now the is gonna be the angle between R and F. Alrighty. So let's think about this. OK? If we have the force, OK. From the sun acting on mercury, that's gonna be in the same direction as RKR is gonna point from the sun to mercury. We just said that the force is also going to do the same. OK. So R and F are gonna be pointing in the same direction right now. If they're pointing in the same direction, the angle between them is just zero. OK. So sign of data is going to be zero even if they were pointing in opposite directions but parallel to each other. OK. We would have a sine of 180 degrees. We would get zero again. OK. So if you're worried about one of them being negative or something that's gonna give a zero. And in this case, they are pointing in the same direction. All right. So sin theta is zero, then our torque is equal to zero. And so we have no torque encountered by mercury due to the gravitational force around the center of the earth. So we look at our answer choices. Option A and B both say yes, there is two. So we know those are incorrect. So we're gonna eliminate those two. Option. C says there's no torque. OK. So that's good. That's what we want. But it says that it's because the gravitational force and the moment arm vector are always perpendicular. OK. Well, we've just shown that the moment arm vector and the gravitational force are in the same direction not perpendicular if they were perpendicular sine of theta would not be zero. We would not have that torque being zero. So the answer is not C either. So we're left with answer D and D is going to be our correct answer. It says that no, all the positions, the net torque acting upon mercury is zero because the gravitational force in the moment our vector are always in the same direction as we've described. And as we drew in our diagram, OK. So we have answer. D being the correct answer. Thanks everyone for watching. I hope this video helped see you in the next one.