ASTRONOMY 9: HISTORY OF COSMOLOGY

Assignment #11--Solutions

2000 March 7

1.

Why was the Newtonian cosmos required to be infinite?

Newton realized that there was a universal law of gravitation: any mass M attracts another mass m with a force F=GMm/r², where r is the distance between them. If you had a finite Newtonian universe, all the matter would be attracted towards the center, and the universe would quickly collapse! Newton avoided this problem by arguing that the universe was infinite so that there was no center and all the gravitational forces would be balanced, assuming the matter is evenly distributed throughout the infinite space.

2.

Astronauts typically orbit the Earth about 100 miles above its surface. Explain carefully why they are ``weightless''. The reason is not that the astronauts are so far away from Earth that the force of gravity is small! (After all, the radius of Earth is 4000 miles, so the gravitational acceleration is still about $(4000^2/4100^2) = 95\%$ as strong as it is on the surface!) The reason for weightlessness in this case is that the astronauts are in free fall, just as if you were descending in an elevator and the cable suddenly snapped. In that case, your fall would be rudely broken when you hit the ground. But the astronauts have a horizontal velocity which just balances their free fall so that they continue to fall towards the Earth, but they never quite hit it.

3.

If Aristotle could drive, he would say that you have to keep your foot on the accelerator because you need a steady force to keep an object in motion. How would Newton respond? Newton's second law, F=ma, says that if a force is applied to an object, it will accelerate. Now let's say we're driving down the road at constant speed. Why do we need to keep applying a force to the car to keep it moving at constant speed? Because there are other forces involved! Friction and air resistance are forces that pull the car in the opposite direction from its motion. So we need to apply just enough force to cancel these out, then the net force on the car will be zero, and it won't accelerate but will continue to move at the same speed. Otherwise, the frictional forces will slow the car down.

4.

Fun with surface gravity (the gravitational acceleration g at the surface of a planet or star).

(a)

The Sun's surface gravity is about 28 times stronger than Earth's. If you were on the Sun, how much would you weigh (lbs)? What would your mass be (kg)? Remember the difference between mass and weight: mass is an inherent property of an object, while weight is a force caused by gravity (which might be stronger or weaker on different planets). So if I weigh 150 pounds on Earth, I would weigh $150\times 28 = 4200$ pounds on the Sun--about two tons! On the other hand, my mass would still be 68 kg no matter where I was.

Note that when you step on to a scale, you are measuring the gravitational force of the Earth pulling on you, your weight. This force is measured in pounds (imperial units) or Newtons (metric units), and would be different on different planets! So, the conversion 1 kg = 2.2 pounds only works on Earth. On a planet like Mars which has a weaker gravitational field, 1 kg would correspond to a smaller force (weight).

(b)

The radius of the Sun is now about $7\times 10^5$ km. In about 5 billion years, it will expand all the way out to the Earth's orbit, becoming a ``red giant'' star! At that point, how much stronger will the Sun's surface gravity be than Earth's? (Hint: you don't need to use the value of G, just use ratios!)

We know that the surface gravitational acceleration is g=GM/r², where M is the mass of the planet or sun, and r is its radius. So we can find the ratio of the surface gravity for the red giant Sun, g_RG, to the surface gravity for the normal Sun, $g_\odot$:

$$\frac{g_{\mathrm{RG}}}{g_\odot} = \frac{GM_\odot/r_{\mathrm{... ...1.5\times 10^8\,\mathrm{km}} \right)^2 = 2.2 \times 10^{-5}.$$

Surface gravity is much weaker on the big new Sun! Now we were given $g_\odot/g = 28$, and so

$$\frac{g_{\mathrm{RG}}}{g} = \frac{g_{\mathrm{RG}}}{g_\odot} ... ...odot}{g} = 28\times (2.2 \times 10^{-5}) = 6 \times 10^{-4}.$$

So even compared to puny little Earth, the surface gravity of the giant Sun is much lower!

(c)

Earth is 81 times more massive than the Moon, and Earth's radius is 3.7 times larger than the Moon's. What is the ratio of Moon's surface gravity to Earth's?

We have

$$\frac{g_\mathrm{moon}}{g_\mathrm{earth}} = \frac{GM_\mathrm{... ... \frac{1}{81} \times 3.7^2 \approx 0.169 \approx \frac{1}{6}.$$

(d)

Why can astronauts on the Moon jump around like superman?

The astronauts are able to jump with a given amount of force using the muscles in their legs. Once they leave the ground, the gravitational force begins to accelerate them downward, slowing their upward rise. Now on the Moon, this acceleration is only 1/6 what it is on Earth, which means that the jumps will last longer and will allow them to jump much higher than on Earth.

5.

Suppose you are sitting on Earth's equator, rotating at constant speed.

(a)

Explain why the rotation of the Earth is causing you to accelerate. (Hint: what is the difference between speed and velocity?)

Remember that acceleration can be a change in speed or a change in the direction of motion. Things moving in a circle are definitely changing the direction of their velocity, even if they are not changing the speed. Velocity is speed plus the direction.

(b)

Compute this acceleration (in meters per second squared, m/s²). (Hint: see assignment #8 for useful numbers!)

An object on the equator is undergoing uniform circular motion as it is carried around by the rotation of the Earth. Now we know that the acceleration for circular motion is a=v²/r. From assignment #8, r = 6400 km $= 6.4\times 10^6$ m, and

$$v = 1670\,\frac{\mathrm{km}}{\mathrm{hr}} \times \left(\frac{... ...rm{m}}{1\,\mathrm{km}}\right) = 460\,\mathrm{m}/\mathrm{s}.$$

Then the acceleration is

$$a = \frac{v^2}{r} = \frac{(460\,\mathrm{m}/\mathrm{s})^2}{6.4\times 10^6\,\mathrm{m}} = 0.033\,\mathrm{m}/\mathrm{s}^2.$$

(c)

How short would a ``day'' (one rotation) have to be in order for you to be in danger of being thrown off into space? Note that the acceleration we just calculated is very small compared to $g\approx 10$ m/s². This means that the Earth would have to be spinning very fast in order to throw you off!

As long as the force of gravity holding you on the Earth is greater than the rotational acceleration, gravity will win and you will be held down. But as we make the velocity higher and higher, eventually we get to the point where things could be ``flung off''. Using F=ma, with the force of gravity mg and the acceleration a=v²/r,

$$mg = \frac{mv^2}{r}.$$

This gives $v=\sqrt{rg}$. Taking the radius of the Earth to be $r=6.4\times 10^6$ m and g=10 m/s², we get v = 8000 m/s. As in assignment #8, the velocity is

$$v = \frac{2\pi r}{t},$$

and so $t = 2\pi r/v \approx 5000$ seconds. The day would be less than 1.5 hours long!

6.

Tides.

(a)

According to Galileo, what did the tides prove?

Galileo thought that his theory of the tides was the proof that the Earth moved--the proof of Copernicanism before a disbelieving Church.

(b)

What was wrong with Galileo's ideas about the tides?

Galileo understood the principle of inertia, and the fact that motion is relative. If you are in a boat with no windows, there is no mechanical experiment you can do onboard the boat to tell that you are sailing downstream (as long as your velocity is not changing, of course). So his theory of the tides is bizarre because it blatantly contradicts his own ideas about inertia! The idea was that by combining orbital and rotational velocities, you get a faster total velocity at midnight and a slower total velocity at noon, so the water gets ``left behind'' (low tide) at midnight and ``rushes ahead'' (high tide) at noon.

Of course, the water doesn't really get left behind, any more than objects dropped from the mast of a moving ship get left behind (they don't!).

An obvious observational problem with the theory was that it predicted only one cycle of high/low tides per day, while it was well known that there were really two.

(c)

It turns out that tidal forces decrease with the cube of the distance: $F_{\mathrm{tidal}} \propto M/r^3$. The Sun is 390 times farther than the Moon, and 27 million times more massive. What is the ratio of tidal forces on Earth due to the Moon and Sun?

The ratio is just

$$\frac{F_{\mathrm{sun}}}{F_{\mathrm{moon}}} = \frac{M_{\math... ...mathrm{sun}}\right)^3 = \frac{2.7\times 10^7}{390^3} = 0.46.$$

(d)

Does the Sun have a significant effect on the tides? The Sun's influence on the tides is about half as great (46%) as the Moon's! So when the Moon, Earth, and Sun line up (new or full moon), the tidal forces are enhanced and the tides are more extreme (``spring'' tide). When the Moon is 90$^\circ$ away from the Sun (``first or third quarter'' Moon), the Sun's tidal force partly cancels the Moon's, and the tides are less extreme (``neap'' tides).

7.

You are floating slowly through space, and notice you are about to collide with a classmate! Fortunately, you are holding your Astro 9 reader. Explain how you might use it to avoid this calamity. Suppose that you throw your reader at your classmate. By Newton's third law, every action has a corresponding reaction. So by pushing the reader away from yourself, the reader has in turn pushed you away with an equal force! Of course, the acceleration of the reader will be much greater than your acceleration since its mass is much smaller. But if you are moving slowly enough, or if you throw it hard enough, then you could reverse direction!

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