🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 22 Easy Difficulty

On earth, two parts of a space probe weigh 11000 $\mathrm{N}$ and 3400 $\mathrm{N}$ . These parts are separated by a center-to-center distance of 12 $\mathrm{m}$ and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

Answer

$1.8 \times 10^{-7} N$

Discussion

You must be signed in to discuss.
KL

Kristel L.

March 21, 2021

The mass of a robot is 5450 kg. This robot weighs 3620 N more on planet A than it does on planet B. Both planets have the same radius of 1.33x10 7 m. What is the difference MA-MB in the masses of these planets?

Video Transcript

we begin discussion by completing the mass off each of this part's So we know that the weight force is given by W is he goes to the mass times the gravity in this case on earth and we know that the gravity on earth is it close to 9.8 meters per second squared. Then the weight force is equal to the mass times 9.8 meters per second squared. Now we can sew for the mass by sending these 9.8 to the other side to get that the mass is it goes to the weight divided by 9.8. Then we used his equation to couple into the wheat off both parts off this probe Number one and number two for the part number one we have m one is the question the wait number one which is 11 3 zeros, new toes divided by 9.8. No, for the other part, we have a mass m to that is it goes to 34 00 divided by 9.8. Now we have moved masses and one and then true. The next thing we have to do is calculate what is the gravitational attraction between both of these parts. How can you do that by remembering that the gravitational force is it goes to new toes, constant times the mass number one times the mass number two divided by the distance between the center's off Each of these masses squared. Then, by using the values that the problem give us, we have that the gravitational force is a coast to G times the mass number one, which is 11 000 divided by 9.8 times 34 00 divided by 9.8 and he's wolfing is divided by R Squared. For this question are squared. Is these 12 meters which is the distance between the center's off both parts, then 12 squared. Now remember that value off new toes constant is that close to 6.67 times 10 to minus 11 new toes times meter squared, divided by kilograms squared. Then the gravitational force is. It goes to 6.6 to 7 times 10 to minus 11 times 11 000 times 34 00 divided buying 12 squared times 9.8 squared And this gives us a gravitational force off approximately 1.8 time stand to minus seven new terms