SAT Physics Subject Test

Chapter 3 Newton’s Laws


Newton’s law of gravitation is really a statement about the force between two point particles: objects that are very small in comparison with the distance between them. However, Newton also proved that a uniform sphere attracts another body as if all of the sphere’s mass were concentrated at its center, so we can also apply Newton’s law of gravitation to objects that are not small, relative to the distance between them.

Therefore, r is the distance between the centers of mass of the two objects. For objects with uniform density, this is merely the distance from center to center.

The force of gravity is very small unless at least one of the objects is large, like a planet or moon. Let’s try a problem that explores this.

  5. The Sun has a mass of 2 × 1030 kg and Mars has a mass of 6 × 1023 kg. How does the acceleration of the Sun due to Mars compare to the acceleration of Mars due to the Sun?

Here’s How to Crack It

The force on the Sun is F = mSaS, and the force on Mars is F = mMaM. Since F is the same in both of these equations—because F is their mutual gravitational attraction—we can write

So, the acceleration of the Sun is much smaller than that of Mars; it’s only 3 × 10–7 times as much. Because of Mars’s much smaller mass, it’s affected more by the gravitational force, which is why Mars orbits the Sun and not the other way around.

  6. If M is the mass of Earth, then the mass of the Moon is about M/80 and the mass of the Sun is about 330,000M. If R is the distance between Earth and the Moon, then the distance between Earth and the Sun is about 400R. So, which exerts a greater gravitational force on the earth: the Moon or the Sun?

Here’s How to Crack It

According to Newton’s law of gravitation, the gravitational forces on Earth exerted by the Moon and by the Sun are:


Simplifying these expressions, we compare the values of


We now see that the gravitational force exerted by the Sun is greater than that exerted by the Moon (by a factor of about  = 160).

  7. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius R around the earth (mass M). What is the speed of the satellite?

Here’s How to Crack It

The centripetal force on the satellite is provided by Earth’s gravitational pull. Therefore

Solving this equation for v yields

v =

Notice that the satellite’s speed doesn’t depend on its mass; even if it were a baseball, if its orbit radius were R, then its orbit speed would still be .

  8. A communications satellite of mass m is orbiting the earth at constant speed in a circular orbit of radius R. If R is increased by a factor of 4, what happens to T, the satellite’s orbit period (the time it takes to complete one orbit)?

Here’s How to Crack It

From question 4, we know that

v =

Now, since v = 2πR/T, we have

This tells us that T is proportional to R, so if R increases by a factor of 4, then T will increase by a factor of 4 = 8.