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As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minimum gravitational force by 11$\% .$ Find the ratio $r_{\max } / r_{\min },$ where $r_{\max }$ is the moon's maximum distance from the center of the planet and $r_{\min }$ is the minimum distance.

1.05

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Numerade Educator

Simon Fraser University

Hope College

McMaster University

So the directory off the moon around the planet is not a perfect circle. It's rather something like on the lips like this one. So in this orbit there are two special distances the maximum distance between the satellite on the planet and the minimum distance between the satellite and the planet. Now, remember the following the gravitational force increases, as did the stents decreases. Therefore, the gravitational force will be bigger in this point then in these points. So here we have the biggest gravitational force. Let me call it F B. While here we have the smallest gravitational force. Let me call it F s. Then the problem says that the Rev additional force F B is 11% bigger than the reputational force. F s should be is equals to 1.11 times f s. Now, what is the relation between the maximum and the minimum distances? So remember the equation for the gravitational force. Magnitude after reputational force is equal student utan, constant time. The man times the mass number one times the mass number two divided by the distance between these masses squared. So what is the reputational force at the point? FB It's equals to the Newtown. Constant times the mass off the Earth times the mass off the moon divided by that minimum distance squared And on the other side what? We have 1.11 times the magnitude off this molesting gravitational force. Which is it? Costa G times the mass off the Earth times the mass off the moon divided by that maximum radios squared. Not that there are a lot off common terms on both sides, so we have a lot of simplifications. G is the same. The minds off the earth is the same. The mask off the moon is the same, so we're left. If the following one divided by the minimal radios squared is equal to 1.11 divided by the maximum radios squared. Then we can send the maximum radios squared to the other side, multiplying to get our marks squared. Divided by arming. Squared is he goes to 1.11 taken the square it off both sides of the equation. We got our much divided by army. Is it close to the square root off 1.11 which gives us a ratio off approximately one 0.5 So the maximum radios this one is only 5% bigger than the minimum radius. These one in the situation

Brazilian Center for Research in Physics