Mars has a mass of 6.4 x 10^23 kg. it has a moon named deimos with a mass of .48 x 10^15 kg.

a. if their centers are 23,460 km apart, what is the force of gravity acting on deimos?

b. a scientific satellite of mass 800 kg orbits mars 400 km ab one its surface. if mars has a radius of 3396 km, what is the force of gravity acting on the scientific satellite?

c. the satellite is in a circular orbit around mars. how does the kinetic energy of the satellite change as it orbits mars?

d. how does the gravitational force on the satellite change as it orbits mars?

PART A)

From gravitational force formula we know that

now we have

now we have

Part B)

again from same formula as we used above to find the force

now we have

now we have

PART C)

here in circular orbit the speed of the satellite will remain the same always

So here kinetic energy of satellite will always remain same

There is no change in kinetic energy of satellite

Part d)

Since satellite is revolving in circular orbit so the distance will always remain the same

As well as the two masses is also constant

So here the gravitational force between them is always same and it will not change

The work done by the professor on the briefcase is zero.

Explanation:

Hi there!

Work is done by a force on an object when the applied force (or one of its components) is parallel to the direction of movement.

The professor applies a force in the briefcase that is perpendicular to the displacement (it points in the vertical direction) that allows him to keep the briefcase in the air. When the professor enters the elevator, it is the elevator that moves him along with the briefcase, so there is no work done by the professor there.

During all the trajectory, the professor did not move the briefcase in the vertical direction except when he picks it up and then deposits it on the floor of the classroom.

When he picks the briefcase, the work done on it is positive (the force is applied in the same direction as the displacement) but when he deposits it in the floor, the work done is negative (the force is applied in opposite direction to the displacement of the briefcase). When he picks the briefcase up, the magnitude of the work done by the professor has to be greater than when he deposits it because the gravity force does work in the down direction that helps him to deposit the briefcase. We can assume that the distance of the briefcase to the floor is small and thus neglect this work.

Then, the net work done by the professor on the briefcase is zero.