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Problem 31 Medium Difficulty

ssm The mass of a robot is 5450 $\mathrm{kg}$ . This robot weighs 3620 $\mathrm{N}$ more $\mathrm{e}$ e on planet $\mathrm{A}$ than it does on planet $\mathrm{B}$ . Both planets have the same radius of $1.33 \times 10^{7} \mathrm{m}$ . What is the difference $M_{\mathrm{A}} M_{\mathrm{B}}$ in the masses of these planets?

Answer

$1.76 \times 10^{24} \mathrm{kg}$

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Video Transcript

in this question, you have to calculate the difference in the masters of planets and and be so be more specific we have to complete one is m A minus MB. So how can he do that? Even this information that we have. So we begin by trying to relate the weight with the mass off the planet. So what is the relation between weight and the mass? Off the planet there is producing the gravity. So remember, that's the weight force is equal to the mass off the object times the acceleration of gravity in the surface off some planet. Now remember that the acceleration of gravity in the surface of a planet is given by the following equation. It's a goes to the Newtown constant times the mass off the planet, divided by the radios off the planet squared. Then we can write the weight force as follows, so the weight force can then be returned. SG times, M times the mass off the planet divided by the radius of the planet squared. This is the well known equation for the gravitational force. But now we will apply these equation using these information. How can you do that? Well, what is the weight off that robot in Planet A? Well, W A is equals to the Newtown Constant. It doesn't change times the mass off the robot times the mass off planet A divided by the radios off Burnett Day. And what is the weight off the rope? What in planet be? Well, very, very similar to the weight on Planet A. But we only have to change the A's by beast so they wait on Planet B is because to Newton, constant times demands off the robot times the mass off planet be divided by the radios off Planet B squared for at this squared here. Then what do we do? We used these equations, so w a g times the mass off the rowboat times the mass of planet B divided by the radios off in it. A squared is ecos too. The value B which is given by the Newton constant times The mass off the robot times the mass off planet me divided by the radios off planet me squared plus 36. 20. No, What we want to couple it is this quantity I m a minus and beat. How can we get this quantity. We have to send this term to the other side of this equation to get the following deep times m r Divided by r A squared times m A minus G times M r times and beef divided by R. V. Squared is he goes to 36 20. And now what now we factor out g m r divided by r squared white Can you do that? Because the radios have planted a easy cause to the radios off limits me so we can factor out these three quantities. So we get the following g m r. Divided by R A, which is equal to R V squared times. I m a minus and be Is it close to 36. 20. Now let me clear the board so I can continue to serve in this equation. Now, we already have the quantity m eight minutes and be right here. So all we have to do is send this term to the other side dividing to get em a minus, and meat is equals to 36 20. Divided by G fines. M r. Divided by r. A. It's weird, and this we can simplify to get the following these Mexico's to 36 20 times are a squared, divided by g times m r. Remember that G The Newton constant is a close to 6.6 to 7 times stand minus 11 new terms meters squared, kilograms squared, then my plague in the vials that were given by the problem. We have the following an A minus and B is equal to 36 20 fines, 1.33 times Stand to the seven squared, divided by 6.6 to 7 times Stan to minus 11 times 54 50 and this is approximately 1.76 times stand to 24 kilograms.

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