# Answers/Solutions: Giancoli Chapter 6

Q1: Yes, always. The force equals GM_{Earth}M_{Apple}/R^{2} where R is the distance between the center of the apple and the center of the Earth. The force of the apple on the Earth is the same as that of the Earth on the apple. For an apple on the Earth's surface, R = R_{E}, the radius of the earth. Then if we define g = GM_{Earth}/R_{E}^{2} = 9.8 m/s^{2}, the two forces are each M_{Apple}g (the directions are opposite, of course). This is true whether the apple is stationary or in free fall. Aside: In the case of free fall, this force is the only force that acts, so M_{Apple}g = M_{Apple}a_{Apple}, which leads to Galileo's result: a_{Apple} = g. It is also true that the Earth accelerates toward the apple: in this case M_{Apple}g = M_{Earth}a_{Earth} which, when solved for the Earth's acceleration, gives a_{Earth} = (M_{Apple}/M_{Earth})g. Since an apple's mass is - for a BIG apple - 1.0 kg and the Earth's mass is about 6x10^{24} kg, the Earth's acceleration is immensely smaller than the apple's.)

Q14: The answer is (b): when the elevator accelerates upward.Your "apparent weight" is the force you exert downward on the scale, which is the same as the normal force the scale exerts upward on you. So there are two vertical forces on you: F_{N}
upward and Mg downward. The vertical equation of motion is, then (with up positive):

F_{N} - Mg = Ma

So, then, your "apparent" weight is

F_{N} = Mg + ma = W + ma

where W is your "true" weight, Mg. If a is positive (upward), then your apparent weight will be greater than your true weight. If a is downward (negative), your apparent weight will be less than your true weight. If a is zero (the elevator stationary or moving with constant velocity) the two will be the same. In free-fall, where a = - g, your apparent weight would be zero.

P5: Let the *original* radius, mass, and g be denoted by the naught sub-script. Then

g_{o} = GM_{o}/R_{o}^{2}

Now if the radius and mass assume new values, R and M, we have

g = GM/R^{2}

So, if M = 2M_{o} and R = 3R_{o}, the new g would be

g = G(2M_{o})/(3R_{o})^{2} = (2/9)GM_{o}/R_{o}^{2} = (2/9)g_{o} = (0.222)g_{o}

P38: Assume a circular lunar orbit. The gravitational force must provide the centripetal force to generate the acceleration toward the Earth:

GM_{Moon}M_{Earth}/D^{2} = M_{Moon}v^{2}/D

where D is the Earth/Moon separation distance (3.84x10^{8} meters, from Giancoli inside cover). The lunar mass divides out. Solving for M_{Earth} we obtain

M_{Earth} = Dv^{2}/G

To get a number for v, use the fact that the Moon orbits the Earth once (traveling a distance of 2D) in a time T = 27.3 days (27.3 days x 24 hours/day x 3600 seconds/hour = 2.36x10^{6} sec) giving a speed of

v = 2D/T

Thus

M_{Earth} = D(2D/T)^{2}/G = (4^{2}D^{3})/(GT^{2})

Using numbers, we arrive at M_{Earth} = 6.02x10^{24} kg.

Comment 1: This is within round-off accuracy of the accepted value of 5.98x10^{24 kg}. One reason we cannot be absolutely accurate using this method is that the Moon's orbit is really an ellipse, not a circle, although the ellipse is very, very close to being a circle.

Comment 2: Look at Question 21, p. 157. In fact, we can use the method of this problem to calculate the mass of ANY astronomical body that has a satellite whose distance from the body and time of orbit is measurable! Using the Earth/Sun data, we obtain the mass of the Sun!

Comment 3: Look at Problem 41, p. 159. Here we apply this method to the Sun's orbit around the center of the Milky Way galaxy. The data here are estimates, but we use this same model. In this case, we assume an estimated value for the galactic mass and then calculate the orbital period, which comes out to be about 200 Million years, sometimes called a "Great Year." Thus, as you may know from other courses, one Great Year ago on Earth the continent of Pangea was breaking up; it was the division between the paleozoic and mesozoic eras.