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Problem 23 Easy Difficulty

A raindrop has a mass of $5.2 \times 10^{-7} \mathrm{kg}$ and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

Answer

$5.1 \times 10^{-6} N$
$5.1 \times 10^{-6} N$

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Video Transcript

the magnitude of the reputational force exerted on the ring drop by the earth is given by the following equation. The gravitational force Is it close to the Newton's constant times The mass off the raindrop times the mass off the earth divided by the distance between the center off the earth on the raindrop squared. Now note the following the radio's off the earth is way bigger than the distance between the raindrop on the surface off the earth. So we say that the distance between the raindrop on the surface off the earth in the situation is negligible, so we do not need to include it in the situation. Then, by substituting your values that were given by the problem, we get the following the gravitational force he's given by 6.6 to 7 find Stand to minus 11 times 5.2 times. Stand to the mind of seven times 5.98 times Stan to 24 and his wolfing is divided by the radios off the earth off 6.38 time Stand to the sixth squared. This gives us a gravitational force with a magnitude off approximately 5.1 time stand to the minus six new terms. Now, for the second item, we have to complete What is the magnitude off the reputation off horse exerted only Earth Bailey, mask off the raindrop Note that the magnitude off this force would also be given by these equations. Sure, these two forces have the same night attitude. The only thing that changes is the direction they're pointing. So the raindrop, uh, pull the earth towards it and the earth pull the raindrop towards it. But if the same magnitude therefore, the magnitude off the gravitational force exerted on Earth by the raindrop is it goes to the magnitude off the gravitational force exerted on the rain drop by the earth, which is what we had just calculated through. Let me right this super screams for clarity. And this is approximately 5.1 time stand to the minus six Newton's. So it's not necessary to do the population again, as the only difference between this forces is their direction.