Rocket Trip Name __Key___

Anne, Bev, Chuck, and Dean have created a space travel scenario involving a rocket that accelerates at 32 ft/s² for 40 years of travel. The trip includes ten years of acceleration and then ten years of deceleration that will bring the rocket to a stop some distance from Earth. One of the objects of this assignment is to find out how far the rocket traveled during this 20 year odyssey.

As soon as the rocket comes to a stop far from Earth, it begins its trip back which is the reverse of the trip away from Earth. Ten years of accelerating toward Earth followed by ten years of decelerating so that the rocket reaches Earth with zero speed.

As mentioned in class, you are going to analyze the trip by breaking it into 3650 short trips each lasting one day of rocket time. Therefore the unit of time for this problem is one day and the unit of distance is 1 light-day, the distance light travels in one day. These definitions insure that the speed of light is one, c = 1 light-day/day.

1. How many seconds in a day? Seconds in a Day = 86,400

The rocket starts each day in a new reference frame that is moving with the rocket. The rocket’s velocity in this frame starts at zero but slowly increases because the rocket is accelerating at 32 ft/s².

2. After one day of accelerating, how fast is the rocket moving in the local reference frame? Speed of Rocket = _2,764,800 = 2.7648 x 10⁶ ft/sec = _1,885,091 mph. (This will be a large velocity compared to the everyday velocities we are used to.)

The speed of light is 1 ft/ns or 1,000,000,000 ft/s = 10⁹ ft/s.

3. After one day of travel in the local reference frame, what is the speed of the rocket, U, as a fraction of the speed of light? U = Speed of Rocket/c = __0.00276_ light-day/day

Note that once the speed is written as a fraction of the speed of light, it doesn’t seem nearly so large! Also notice that the speed has the same numerical value in ft/ns, light-days/day, or light-years/year. That is the beauty of using velocities as fractions of light speed.

The U found in question 3 is the value that will be used in the addition of velocity formula in the table that you will create for this assignment.

4. Suppose we did not know anything about Special Relativity and used the answer in question 3 to find the speed of the rocket after 3650 days of accelerating. What would we calculate for the rocket’s velocity after those 3650 days of constant acceleration?

5. The Velocity of the Rocket Ignoring Special Relativity = 10.3 times the speed of light.

6. Any comments about your answer to question 5?

Wow. Too bad Special Relativity does not allow the rocket to go that fast!!

7. Create an Excel spreadsheet with the following columns:

Trocket

Tearth

Xearth

Vstart

Vend

Vaverage

ΔTearth

ΔXsam

a. T_rocket starts at 0 and increases 1 day per row. The first row corresponds to the rocket’s trip during day one. The second row corresponds to the rocket’s trip during day two, etc.

b. T_earth starts at zero and increases by ΔT_earth each row.

c. X_earth starts at zero and increases by ΔX_earth each row.

d. V_start is the velocity of the rocket with respect to Earth at the start of each day’s trip. It is equal to zero in the first row. In each subsequent row, it is equal to the V_end from the previous row.

e. V_end = is the velocity of the rocket with respect to earth at the end of each day. We learned earlier in the class that the reasonable and obvious way to add velocities, V_end =, is incorrect for velocities that are large compared to the speed of light. The U in the equation is your answer to question 3, the rocket’s speed in the local reference frame after accelerating for 1 day.

f. V_average =

g. ΔT_earth = ΔT_rocket / but since ΔT_rocket = 1 day, ΔT_earth = 1 / days

h. ΔX_earth = V_average ΔT_earth

8. Each new row is found by using the values in the previous row. The table gets built one row at a time. Excel is very content to do the same calculations over and over again, row by row, for as many rows as you want!

The tables below are there to help you get your table straight before having Excel do the calculation for 3650 rows.

Trocket	Tearth	Xearth	Vstart	Vend	Vaverage	ΔTearth	ΔXearth
0	0	0	0	0.0028	0.0014	1	0.0014
1	1	0.0014	0.0028	0.005599956	0.004199978	1.000004	0.0042
2	2.000004	0.0056	0.005599956	0.008399824	0.00699989	1.000016	0.007
3	3.00002	0.0126	0.008399824	0.011199561	0.009799693	1.000035	0.0098

Trocket	Tearth	Xearth	Vstart	Vend	Vaverage	ΔTearth	ΔXearth
0	0	0	0	=+(D2+0.0028)/(1+D2*0.0028)	=0.5*(D2+E2)	=+(1-D2^2)^(-0.5)	=+F2*G2
=+A2+1	=+B2+G2	=+C2+H2	=+E2	=+(D3+0.0028)/(1+D3*0.0028)	=0.5*(D3+E3)	=+(1-D3^2)^(-0.5)	=+F3*G3
=+A3+1	=+B3+G3	=+C3+H3	=+E3	=+(D4+0.0028)/(1+D4*0.0028)	=0.5*(D4+E4)	=+(1-D4^2)^(-0.5)	=+F4*G4
=+A4+1	=+B4+G4	=+C4+H4	=+E4	=+(D5+0.0028)/(1+D5*0.0028)	=0.5*(D5+E5)	=+(1-D5^2)^(-0.5)	=+F5*G5
=+A5+1	=+B5+G5	=+C5+H5	=+E5	=+(D6+0.0028)/(1+D6*0.0028)	=0.5*(D6+E6)	=+(1-D6^2)^(-0.5)	=+F6*G6

9. After ten years of travel, or 3650 days, the rocket’s distance from earth is _4,290,596_ light-days or 11,755_ light-years.

10. After ten years of travel, the earth-frame observer next to the rocket has a clock that reads _4,290,958_ days or _11,756_ years.

11. Earth observers, using your answers to questions 7 and 8 to calculate the rocket’s average velocity during the first ten years, get the average velocity to be _0.99991 or effectively 1.000 _.

12. When rocket returns to Earth, the maximum distance from Earth was 23,510 light-years.

13. Upon the rocket’s triumphant return, the Earth clock that read zero at the start of the trip now reads _47,010_ years.

14. How much Earth time elapsed during the 3650^th day of the rocket’s trip? (The rocket’s speed is as big as it ever will be at the end of this day.) _11,859_ days or _32.5_ years!

15. How does the rocket’s maximum distance compare to size of the Milky Way, our home galaxy? About ¼ of the diameter of the Milky Way.

16. What does that say about the technological feasibility of visiting other galaxies?

Chances run between none and slim but slim depends on finding some way to overcome the limitations established by the nature of space and time.