PTYS206-2 Spring 2008 Homework #6 KEY. a) Acceleration due to gravity: a = GM/R 2 If radius is doubled, does acceleration increase or decrease?

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1 PTYS206-2 Spring 2008 Homework #6 KEY 1. Understanding physical relationships (part 1). a) Acceleration due to gravity: a = GM/R 2 If radius is doubled, does acceleration increase or decrease? b) Escape velocity: V esc = (2GM/R) 1/2 If mass is doubled, does escape velocity increase or decrease? c) Newtonian formulation of Kepler s third law: P 2 = (4π 2 /GM)R 3 If radius is reduced by 1/2 does P 2 increase or decrease? If mass and radius are doubled, does P 2 increase or decrease? d) Position under constant acceleration: x = ½ at 2 If time is doubled, does position increase or decrease? If time is doubled and acceleration is reduced by ½, does position increase or decrease? e) Wave speed: c = λ/t If wavelength is reduced by ½, does wave speed increase or decrease? If wavelength and frequency are both doubled, does wave speed increase or decrease? The frequency of the wave (call it f) is related to the period of the wave as follows: f = 1/T.

2 So we can rewrite the equation for the wave speed c as: c = λ / T = λ f. Using this new equation, we can see that if we double the wavelength and the frequency, the wave speed will go up by 4. Since the original assignment mistakenly did not include the new equation relating wave speed to frequency, students who used the original equation and answered that the wave speed wouldn t change got full credit. Students who figured out the new equation and gave the correct answer got an extra point. 2. Understanding physical relationships (part 2). Looking back at each part in problem 1, determine numerically how the calculated parameter changes in the given situations. For example, in 1a) R is doubled, how much does a change by? Show your work using an additional page if necessary. There are two methods you can use to solve these problems. First, you can choose sample values for each parameter in the equation and plug in those values to figure out how the quantity in question changes. This method has been used for the first two questions: a) Acceleration due to gravity: a = GM/R 2 Since we are only interested in how the acceleration changes when we change R and keep G and M the same, we can let G, M and R = 1. In this case: a 1 = 1*1/(1 2 ) = 1 (units are not relevant here) Now, the problem asks us to double R, while leaving the other parameters the same. Thus we have G and M = 1, while R = 2. Now: a 2 = 1*1/(2 2 ) = ¼ Comparing the two values gives: a 2 /a 1 = (1/4)/1 = ¼. So, when R is doubled, a is reduced by a factor of ¼. b) Escape velocity: V esc,1 = (2GM/R) 1/2 Like in part a), we will set G, M and R = 1. This gives:

3 V esc,2 = (2*1*1/1) 1/2 = (2) 1/2 We are asked to double M. Thus we have G and R = 1 and M = 2. So: V esc,2 = (2*1*2/1) 1/2 = (4) 1/2 = 2 = [2 1/2 ]*[2 1/2 ] Again, comparing the two values gives: V esc,2 /V esc,1 = [2 1/2 ]*[2 1/2 ]/ (2) 1/2 = (2) 1/2 So, when M is doubled, V esc is increased by a factor of 2 1/2 As noted above, you can use this method to determine physical relationships from equations. However, it can be a bit tedious. We can also use algebra to determine how the prescribed changes affect a quantity. By substitution and rearranging of the equations, the relationships can be determined. c) Newtonian formulation of Kepler s third law: P 2 = (4π 2 /GM)R 3 For the first situation we want to know what happens to P 2 if R is reduced by ½. We will define a new variable, R. R = ½ R. x = (4π 2 /GM)(R ) 3 = (4π 2 /GM)(1/2*R) 3 = (1/2) 3 (4π 2 /GM)R 3 We ve rearranged this equation to remove the numeric factor and leave all of the terms from the definition of P 2. From above, we can substitute the definition of P 2 and see directly what happens with the given change: x = (1/2) 3 * P 2 = (1/8)*P 2 Thus, when R is reduced by 1/2, P 2 is reduced by a factor of 1/8. For the second situation, we will be doubling both R and M so we need to define two variables: M = 2M and R = 2R Thus:

4 x = (4π 2 /GM )R 3 = (4π 2 /G(2M))(2R) 3 = (2 3 /2)*(4π 2 /GM)R 3 As before, let s use the definition of P 2 : x = (2 3 /2) * P 2 = (8/2)*P 2 = 4*P 2 Thus, when M and R are both doubled, P 2 is increased by a factor of 4. d) Position under constant acceleration: x = ½ at 2 Again, we have two scenarios to examine. For the first situation, we want to know what happens to x if we are double t. Thus we need to define one new variable: t = 2t Rather than using x, as we did in the last problem we will use z, just to avoid confusion: z = ½ a t 2 = ½ a (2t) 2 = (2 2 ) ½ at 2 Substituting the definition for x gives: z = (2 2 ) * x = 4 * x Thus, when t is doubled, x increases by a factor of 4. In the second scenario, t is doubled and a is reduced by 1/2. Again we need two variables: t' = 2t and a = ½ a So: z = ½ a t 2 = ½ (1/2a) (2t) 2 = (1/2)*(2 2 )* 1/2at 2 Substituting the definition of x gives: z = (1/2)*4*x = 2*x Thus, when t is doubled and a is reduced by ½, x is increased by a factor of 2.

5 e) Wave speed: c = λ/t As with the last two problems, we have two different scenarios to examine. In the first case, λ is reduced by ½. λ = (½)λ So for this case: x = λ /T = (1/2) λ/t Substituting the definition of c gives: x = (1/2)*c Thus, when λ is reduced by ½, c is also reduced by ½. For the final scenario, both λ and T are doubled. Defining the two necessary variables gives: λ' = 2 λ and T = 2T Now we have: x = λ /T = (2λ)/(2T) = (2/2) λ/t When we substitute the definition of c, we have: x = (2/2)*c = 1*c = c Thus, when both λ and T are doubled, c stays the same! For those students who caught the typo, the appropriate formula was not x = λ /T, but rather x = x = λ * ν = 4 * c. The goal of these kinds of problems is to understand the physical relationships described by an equation. It is not enough just to memorize the equation. To understand the science behind the mathematics, one must be able to recognize the relationships between the physical parameters.

6 3. Determine the SI units of G using algebra and any relevant formula you choose. Starting with the formula, assign units to all the quantities but G. Then use algebra to determine the units of G. Any equation that involves G (the gravitational constant) can be used here. For illustration, two different equations will be used in this key. Regardless of which equation you choose, the first step should be solving that equation for G. Then, assign proper SI units to each of the other parameters and work through the algebra until the solution is in the correct units. A) Using the formula for the acceleration due to gravity: a = GM/R 2 Solving this for G gives: G = ar 2 /M The SI units for each parameter are: acceleration (a): meters per second squared (m/s 2 ) radius (R): meters (m) mass (M): kilograms (kg) Putting these units into the formula for G gives: G = (m/s 2 )*(m) 2 / kg = (m/s 2 )*m 2 /kg = (m 3 /s 2 )/kg = m 3 /s 2 kg B) Using the formula for escape velocity: V esc = (2GM/R) 1/2 Solving for G gives: G = (V esc ) 2 *R/ (2M) The SI units for each parameter are: escape velocity (V esc ): meters per second (m/s) radius (R): meters (m) mass (M): kilograms (kg) The constant 2 has no units and MUST be ignored here.

7 Putting these units into the formula for G gives: G = (m/s) 2 * (m)/ kg = (m 2 /s 2 )*m/kg = (m 3 /s 2 )/kg = m 3 /s 2 kg As expected, both methods result in the same solution! 4. The visible spectrum is from about 380 nm to 780 nm. What is the frequency range of visible light? We need to determine the range of frequencies represented by this range of wavelengths. The relevant equation is the definition of wave speed: c = λ/t However, this equation compares wave speed (c), wavelength (λ) and wave period (T). We are given wave frequency, not period. Thus we need to make a substitution using the relationship between period and frequency: T = 1/ ν (where ν is the frequency) Thus, the relationship we need is: u = λ/t = λ(1/ν) = λ*ν Solving this for frequency gives: ν = u/λ Since we are talking about the visible spectrum, we are discussing waves that travel at the speed of light. Thus, c = x10 8 m/s The last step we need to do before we can calculate the range of frequencies is to convert our wavelengths into SI units. The conversion factor we will use is: 1 m = 10 9 nm. λ 1 = 380 nm * (1m/10 9 nm) = (3.8x10 2 nm)*(1m/10 9 nm) = 3.8x10-7 m λ 2 = 780 nm * (1m/10 9 nm) = (7.8x10 2 nm)*(1m/10 9 nm) = 7.8x10-7 m

8 Onto the calculations: ν 1 = u/λ 1 = (2.9979x10 8 m/s)/ 3.8x10-7 m = x10 14 /s = 7.89x10 14 Hz ν 2 = u/λ 2 = (2.9979x10 8 m/s)/ 7.8x10-7 m = x10 14 /s = 3.84x10 14 Hz NOTE: The largest frequency is not associated with the largest wavelength, but rather with the shortest wavelength! For light waves, as you increase the wavelength, you decrease the period. The range of frequencies of visible light is thus: 3.84x10 14 Hz to 7.89x10 14 Hz 5. Convert the following to binary numbers. a b c d The solutions are laid out in the following table: 2 7 = = = = = = = =

9 6. Convert the following ascii code into English, i.e. find the letter that corresponds to each binary number in the ascii code: ASTRONOMY (all in caps!)