Kepler's third law and multiple formulas

[stressedout]

Hi,

I just got my physics test back and am hoping I can be helped with two questions. My instructor gave it back at the end of class leaving no time for going over it.

1) Suppose the period of the moon in its orbit about the Earth were 41.1 days and the moon were 768,000 km from earth. Suppose the Earth's radius were 15,950 km. Use Kepler's third law to determine what the period of an artificial satellite orbiting near the Earth's surface would be.

Kepler's third law is: T^2=Kd^3 or from what I understand D^3/T^2
Tm= 41.1 days
Dm= 768,000 km
De= 15,950 km

So first I did:

Dm/De= 768,000 km/15,950 km = 48.15047022 km

Then (Dm/De)^3= 48.15047022^3 km = 111635.3139 km

Next I did (Tm/Te)^2= 111635.3139 km
so the square root of 111632 = 334.1187123

334.1187123 x 41.1 days = 13732.27908 days
now I have to get in into hours so:
13732.27908 days x 24 hr/day = 329574.6978

Now what? Have I done something wrong or do I have more steps to take because the answer should be one of these four choices:

(a)1 hr (b) 2 hr (c) 3 hr (d) 4hr (e) 5 hr


2) Calculate the carrying capacity (number of cars passing a given point per minute) on a highway with three lanes (in one direction) using the following assumptions: The average speed is 100 km/hr, the average length of a car is 7.0 m, and the average distance between cars should be 75 m.

v= 100 km/hr
d= 75 m

I thought I shoud begin the problem by looking for the time it takes so I did t=d/v and got t= 0.75 s. Now I am stuck. I know I have to use another formula somewhere. Please let me know what I am looking for now and which formula I should use so I can try to get the answer.

The answer needs to include cars/min as the choice for the answers are:

(a) 21 cars/min (b) 31 cars/min (c) 41 cars/min (d) 51 cars/min
(e) 61 cars/min


Is there any type of clue you can give me to know when to use which formula when answering physics question. Unless I see certain words or symbols mentioned like centripetal acceleration, average acceleration, or N in questions I am at a loss. There are so many formulas and I am not clear on when which formula should be used.

Thank you in advance.

 

[Janus]

Actually, Kepler's third, or "Harmonic" law is:

T₁²/T₂²=D₁³/D₂³

Which relates the orbits of two object, revolving around the same body.

You are given T₁ andD₁, the Moon's period and distance, and D₂, the satellite distance, so all you need to do is rearrange to find T₂

 

[HallsofIvy]

Janus: That was exactly what he did!

1)
stressedout: You were fine up to here:

Next I did (Tm/Te)^2= 111635.3139 km
so the square root of 111632 = 334.1187123

334.1187123 x 41.1 days = 13732.27908 days

You are given Tm and are looking for Te:
it's Tm/Te= 334.12 so Te= Tm/334.12.

You went the wrong way!


2) Since the average speed of the cars is 100 km/h, in one minute, the average car will have gone 100 km/h* 1/60 hr= 1.666667 km or
1667 m. I.e. you will have observed all the cars pass which were up to 1667 m back when you started timing. Since you are told that the average car is 7 m and there is an average of 75 m between cars, each car is taking a total of 82 m of road. There are 1667/82= 20.3 such blocks in 1667 m of road so you should observe an average of 21 (since 20 is not given as an answer- I guess they are counting part of a car as a car) cars per minute.

The way you tried would give you the answer except for two small errors and one horrendous mistake:
The horrendous mistake: 75 m by 100 km/hr is NOT "0.75 S".
100 km/h = 100 km/hr*1000 m/km= 100000 m/h= 100000 m/h *1 h/3600 s= 27.8 m/s so 75 m by 27.8 m/s= 2.7 s. Each car would take 2.7 s to pass, not .75s.

The small errors: first, you should have calculated the time in minutes, not seconds, since you are asked for the number of cars that pass in one minute: 2.7 s= 2.7/60= 0.45 min. So that in one minute
1/.45= 22.2 cars pass.

That's not quite correct because, to find the section of road each car "occupies", you need to include the car itself! You should use 75+ 7= 82 m. for each car. 82/27.8= 2.85 s = 0.49 m and 1/.49 m/car
= 2.04 cars close to what I had.