Quick and Easy Math

How many math ‘word problems’ have you done in your life? You don’t have to answer that, because you will do many more in the future. In the near future, a few of those will be in astronomy.

Let’s start out with a non-astronomy one to give you an idea of what the format will be.

Q1 The cost of ingredients for a 6-inch pizza is 2 dollars. How much do ingredients cost for an 18-inch pizza? Answer

Q2 The amount of light that a telescope can gather (light gathering power) depends on the area of its primary mirror (or lens). The size of a telescope mirror is stated as the diameter. For example UGA has a 24-inch (0.6-meter) telescope. The largest telescope in the USSR is 6.0 meters. How much greater or smaller is the USSR’s telescope’s light gathering power over UGA’s? Answer

The above two examples depend on a simple geometric relationship: the area of a circle is pi times the radius squared [pi x r²], the important part being the radius squared. In the course we will use a few geometric relationships and a few physical relationships to express quantitative behavior of physical systems. As shorthand to remembering how the relationship works we will use formulas. Since the formulas will be on the tests you will not have to remember them, but you will have to remember what geometry or physics they stand for, otherwise you will not know which to use in your calculations. If you have trouble with moving from algebraic symbols to measured quantities and back, I suggest you rewrite the formulas in words. Then spend some time trying to visually imagining what happens when each variable in the equations changes. The applets linked from the lecture notes are helpful in this.

To review – The first thing you need to do to solve a word² problem in to understand and visualize what each of the quantities in the problem is. The second thing to do is to ask what relationship (if any) they have between them. In the problems above if you did not know what diameter or area were, or you did not understand the relationships between them you could not do the problem!

Q3 The frequency of a photon increases by a factor of 5. How much more energy will it have? [Energy of a photon = a constant x frequency; shorthand E =h f ] Answer

Q4 If the frequency of a photon increases by a factor of 5. What will the wavelength of the photon be? [Speed of light = frequency x wavelength; c =f l] Answer

The two questions above involve variables to only the first power. That is, if one thing gets larger (or smaller) then another thing gets larger (or smaller) by the same factor. For example: If it took me 60 minutes to drive to Atlanta at 70 miles per hour, then at half (1/2) the speed = 35 miles per hour, you would get there in 120 minutes. Speed = distance/time or time = distance /speed. So the factor time changes is =1/(1/2) = 2 = twice as long.

Gravity works in this simple way SOMETIMES; as long as you don’t change the distance between the two attractive objects.

The force of Gravity F = G Mm/d² where M and M are the two masses and d is the distance between them. If you increase (decrease) either mass by a factor then the force increases (decreases) by the same factor.

Q5 The Mass of the Sun doubles while the mass of the Earth and the distance between them remains the same. By what factor did the Earth-Sun force change? Answer

BUT the gravity formula has four variables so you can change one, two or three of them before solving for the last one.

Q6 The sun’s mass doubles and the Earth’s mass triples while the distance between them remains the same. How much does the force between them change? Answer

Gravity also has a variable that changes as the square (just as the pizza area above) only the force of gravity gets smaller as the square of the distance. 1/d²

Q7 The Sun and the Earth’s masses stay the same but the distance between the Sun and Earth gets three times larger. How much does the force between the Earth and the Sun change? Answer

Escape velocity has another power. Instead of squared it uses the square root (which can also be noted as ½ power). Thus SQRT (4) =4^1/2 =2.

V_esc = SQRT (2GM/R)

Since 2 and G are constants the escape velocity depends only on the mass (M) divided by the distance to the center of the mass (R), which is usually the radius of the object of mass M.

Q8 The mass of a planet doubles while the size of the planet shrinks by a factor of 2. What is the change in its escape velocity? Answer

A final example; Kepler’s third law. It poses two problems. The first is that it uses a cube so let me remind you that

Z³ =Z x Z x Z

The second problem is that is has an addition in it. This requires you to assess whether the addition is going to make any noticeable difference in the final outcome. The idea is that if you add two masses M₁ + M_2, then if one of them is quite a bit larger than the other, the sum is almost the same number as the larger one by itself. Thus you can set the smaller one to zero. So if the two masses are the Sun and a planet, the sum is very much the same as the Sun by itself. This does not work if you are dealing with objects of similar mass such as two stars.

Q9 The Earth’s mass doubles and the earth-sun distance increases by a factor of 4. How long does it take to go around the Sun? Answer