1105

• Run Multiple Choice Generator at h:\shares\physics
• General College Physics:  Run Core Problems by giving the program 174 as your 'course'--enter this when asked for your 3-digit course number.   Do the problems for 11/03.
• University Physics:  Run Multiple Choice Generator for 201 and choose 'Calculus of Uniformly Accelerated Motion'.
• If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution.

`l If a magnet is positioned at the origin of the x axis, and a steel ball is positioned to the left of the origin, then the force exerted on the ball is to the right. If the ball is moved closer to the origin then it is closer to the magnet and experiences a greater force to the right. The displacement of the ball is to the right, and the forces exerted on the ball are to the right, so the magnetic force and the displacement are in the same direction. The magnetic force therefore does positive work on the ball.{}{}If a magnet is positioned at the origin of the x axis, and a steel ball is positioned to the right of the origin, then the force exerted on the ball is to the left. If the ball is moved closer to the origin then it is closer to the magnet and experiences a greater force to the left. The displacement of the ball is to the left (i.e., in the negative direction), and the forces exerted on the ball are to the left (also in the negative direction), so the magnetic force and the displacement are in the same direction. The magnetic force therefore does positive work on the ball.

The total gravitational-field flux of the Earth is, as we saw last time, about 5 * 10^15 (m/s^2) * m^2.

What is the gravitational field strength of the Earth at the distance of the Moon, which is about 400,000 km away?

At 400,000 km the flux is spread over a sphere of radius 4 * 10^8 meters.  The area of this sphere is about 2 * 10^18 m^s, so the field is

• field strength = flux / area =

5 * 10^15 m/s^2 * m^2 / (2 * 10^18 m^2) =

2.5 * 10^-3 m/s^2.

The total gravitational flux of the Earth is 4 pi G M, where G is the universal gravitational constant 6.67* 10^-11 N m^2 / kg^2 and M is the mass of the Earth.  What is the mass of the Earth?

The total gravitational flux is 5 * 10^15 m^3 / s^2. 

The total gravitational flux is 4 pi G M. 

So

• M = total gravitational flux / 4 pi G

= 5 * 10^15 m^3 / s^2 / (4 pi * 6.67 * 10^-11 N m^2 / kg^2)

=  6 * 10^24   (m^3 / s^2) / (N m^2 / kg^2)

=  6 * 10^24 kg     

(be sure you can work out the units.  Remember that a N is a kg m/s^2).

All this is equivalent to the following:

Newton's Law of gravitation says that two point masses m1 and m2 separated by distance r exert a mutual gravitational force

• F = G m1 m2 / r^2,

with the force directed along the line between the two masses.

The mass of the Earth is 6 * 10^24 kg and you are 6400 km from its center.  Outside the surface of the Earth it acts just as if it was a point mass.  Compared to the 6400 km distance, you are effectively concentrated at a point.  What is the force exerted between you and the Earth?  G = 6.67 * 10^-11 N m^2 / kg^2.

Using mass 75 kg (my mass) we get

• F = G m1 m2 / r^2

= 6.67 * 10^-11 N m^2 / kg^2 * (6*10^24 kg) * (75 kg) / (6.4 * 10^6 m)^2

= 740 N, approx..

An object moving at constant speed v on a circle of radius r is accelerating at rate

• a_centripetal = v^2 / r

in the direction of the center of the circle.

Suppose we want to orbit an object at a 600 km 'altitude'.  We figured out yesterday that the orbital radius would by 7000 km and the acceleration of gravity would be about 8.3 m/s^2 toward the center of the Earth.  How fast would the object have to be moving in order for the centripetal acceleration to be 8.3 m/s^2?

We know that a_centripetal = v^2 / r.  We know that for this situation r = 7000 km = 7 * 10^6 m, and a_centripetal must be 8.3 m/s^2.  So we solve for v, obtaining

• v = sqrt(a_centripetal * r)

= sqrt(8.3 m/s^2 * (7 * 10^6 m) )

= 7600 m/s.