Physics 150 Fall 2003

Solution to HW 2

1. There can be many answers. Possible ideas are:

a. One observes the sun, moon, and stars moving but do not sense that Earth is moving

b. There is really no direct evidence. (Later Aristarchus
of Samos measured the size of the sun to be much
larger than that of the Earth suggesting its greater importance in the heavens
than that of the Earth.) But it could be argued that the sun is more powerful
in some way and therefore at the center.
Indeed the sun
god was the supreme god in

c. The known retrograde motion of certain “stars” (planets) and the varying orbits of the Sun and Moon might suggest that there is no center. This would appear to be a more philosophical argument.

Chapter 2

Question C: Galileo invoked two principles in order to show that the Earth could be moving without us knowing it. The first was inertia: he knew when he rolled a ball off of a table that, after it left his hand, it continues to move at the same speed “away from him” (in the horizontal direction) throughout its entire path, even while it is falling. That meant that if a person jumps, even though the Earth is moving, the jumper will not notice the motion of the Earth under herself because she is moving at the same speed with the Earth. The second principle was superposition: the ball that he rolled off the table continued to roll at the same speed (in the horizontal direction) even after the ball starts falling. Gravity acts in the vertical direction and thus doesn’t affect the horizontal motion of the ball. Therefore, the two motions (horizontal and vertical) are independent and one can “superimpose” the two independent motions in order to observe the real motion of the ball. This discovery implied that gravity would act exactly the same whether the Earth was still or in motion.

**Note: Show all work
on numerical problems. Do not write a
numerical answer only.**

Problem 2

- The
horizontal velocity remains 30 m/s both on and off the cliff for the
entire flight. The car travels this
speed for a time of 2 s before hitting the ground below. Therefore, the total horizontal distance
traveled is 30 m/s * 2 s =
**60 m**.

**The car lands 60 m from the base of the cliff.** - When
the car flies off the cliff it has no initial vertical velocity, v
_{y}^{0}=0. Therefore, y = v_{y}^{0 }* t + 0.5 * a * t^{2}= 0 – 0.5*9.8*2^{2 }=**19.6 m.**

**The cliff is 19.6 m high.**

(if you use a = 10, the answer is**20.0 m**which is OK.)

Problem 4

- The
total horizontal displacement is x = 54 m.
The ball moves at a constant horizontal velocity for the entire
time of t = 6s. Therefore, the
horizontal velocity is v = x/t = 54/6 =
**9 m/s**. - In the
vertical direction the ball starts at speed v
_{y}^{0}going upward; it then decelerates due to gravity (downward acceleration) and has zero vertical velocity when it reaches the highest point, which happens at half the entire flight time, t=3s. Thus 0 = v_{y}^{0}– 9.8*3 or vy0 =**29.4 m/s**. (or**30. 0 m/s**if a = 10) - The
y-position (as a function of time) is given by the formula: y(t) = v
_{y}^{0}*t + 0.5*a*t^2. We want to know: what is the y-position when t = 3 s, that’s when the ball is at its highest. y(t=3s) = 29.4*3 – 0.5*9.8*3^2 =**44.1 m**. (or**45.0 m**)

Chapter 4, Question B:

To answer this question, we need to find information available at the time that would give the ratio of the mass of the sun to the mass of the earth. Two pieces of information can be used to get information on the mass of the earth:

1) the
acceleration of gravity ( F = GmM_{e}/R_{e}^{2} = mg or
g = GM_{e}/R^{2} or GM_{e} =
g R_{e}^{2}, where M_{e} = mass of the earth and R_{e}
= radius of the earth).

2) The period of the moon is given by F = GmM/R_{em}^{2} = ma
= m v^{2}/R_{em}

= m (2pi R_{em} /P_{m})^{2}/R_{em} or GM_{e} = (2pi)^{2 } R_{em}^{3} /P_{m}^{2}
where R_{em}^{3} is the earth-moon distance and P_{m}
is the period of the moon’s orbit (28 days).

Similarly, to point 2) above the mass of the sun is given by
GM_{s} = (2pi)^{2}^{ } R_{es}^{3} /P_{e}^{2}
where R_{es}^{3} is the earth-sun distance and P_{e} is the period of the earth’s motion (365
days)

Thus the ratio GM_{s} /GM_{e}
= M_{s} /M_{e} = (R_{es}^{3} /P_{e}^{2} )/(R_{em}^{3} /P_{m}^{2})
where all the quantities on the right are known. A formula can also be found in terms of g.

Homework problem #3

The picture drawn shows a right triangle. This means that we can use trigonometry to
calculate the distance from the Earth to the moon, d. This distance is the hypotenuse of the right
triangle. The Earth’s radius r is the
adjacent side to the angle L. Using the
trick: SOH-CAH-TOA (sine equals opposite over hypotenuse – cosine equals
adjacent over hypotenuse – tangent equals opposite over adjacent), we see that
we want to use the cosine of L. cos L = adj/hyp
= r/d or d = r/cos L.
We use the approximate radius of the Earth to be r = 3900 miles then d =
3900/cos(89.1) = **248,300
miles**.

Homework problem #4

We now have two right triangles. If we isolate one triangle then it has a
small angle of 0.5*A=1/4 degree. The
hypotenuse is again the distance from the Earth to the moon (subtracted by the
Earth’s radius this time except because the distance to the moon is much larger
than the Earth’s radius we will just use the value from problem 1) Using trig again we see that sin(0.5*A) = opp/hyp = moon’s radius/distance Earth to moon. Or moon’s radius = 238,000 miles*sin(0.25) = **1040
miles**.

Homework problem #5

There are three components to the accuracy of this measurement. 1) the accuracy in knowing the Earth’s radius 2) the accuracy in knowing the values of trigonometric functions 3) the accuracy in measuring the angles above.

Errors in the Earth’s radius are affected by the Earth not being perfectly spherical. This kind of error is not too serious because our answer depends linearly on its measurement, e.g. a 1% error in the Earth’s radius causes a 1% error in the distance. For example, let’s say we used 3939 miles for the Earth’s radius (1% difference from the value we used) then our answer would have come out as around 251,000 miles which is roughly a 1% change from our original answer.

The second type of error is also not too serious. For example, my calculator shows that the value of cos(89.1) = 0.01570. The Ancients would not have had this convenience and would instead have used tables constructed by physically measuring ratios from triangles. If we imagine a 1% error in this valu, our answer is still within 1% of our original answer.

However, the third type of error can be disastrous. For example, if we measured the angle to be 89.95 degrees rather than 89.1, which is less than 1% difference in the angle, then our answer would have been 4.77 million miles which is a 1900% change.