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Problem 21 Easy Difficulty

Mars has a mass of $6.46 \times 10^{23} \mathrm{kg}$ and a radius of $3.39 \times 10^{6} \mathrm{m} .$ (a) What is the acceleration due to gravity on Mars? (b) How much would a $65-\mathrm{kg}$ person weigh on this planet?

Answer

3.75$m / s^{2}$
$2.4 \times 10^{2} N$

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Video Transcript

we know that the weight force that acts on something nearby the surface off a massive body like a planet is given by W is he goes to the mass off that object times acceleration off gravity at the surface off this massive body. On the other hand, we know that the gravitational force is given by F G, which is it goes to biggie times the mass off the massive body, the mass off the planet off the star off the comet off this massive body that that is producing gravity kinds The mess off the object that's nearby, the surface divided by r squared. This R squared is the distance between the object and the center off the massive body. For instance, we have here on the left the surface off planet Mars. If we have an object that is hovering right here, these little are on the F G equation. Would be these distance the distance between the center off planet Mars and the body that is under the effect off its gravity. Now, by comparing these equation with this equation, we see that we can identify this quantity as being the acceleration of gravity at the surface off some massive body. In our case, we want to complete the surface gravity off planet Mars. So using this equation that we have just arrived to sort off, we have d is it goes to biggie times m divided by r squared. Then they plaguing the known values we got G times 6.46 time stand to 23 divided by three 0.39 times 10 to the sixth squared. Now what is the value off g? This is the value off noodles constant. And this is approximately 6.6 to 7 time Stan to minus 11 Newtons meters squared, divided by kilograms squared. Then we have the following surface gravity for planet Mars. The surface gravity is he goes to 6.67 time stand to mine is 11 times 6.46 times stand to 23 divided by three points 39 times 10 to the sixth squared and this is approximately three 0.75 meters per second squared. The next item there is a person off mass M equals 2 65 kilograms on the surface off Mars. How much will disperse? Um wait well to answer this question. We have to use these lives for decorative acceleration of gravity with this equation for the weight force, then we have the following the weight off that person is it close to 65 kilograms times tree 0.75 meters per second squared and these is approximately 2.4 times 10 squared new tones.