couple of q's (1 Viewer)

[clever angel]

hi

can someone answer these asap

an observer in a closed lab wishes to demonstrate whether he is at rest or constant velocity. which is correct?

a. he can find out by measuring the apparent velocity of light in lab
b. he cannot find out
c. he can find out by comapring two different clocks in the lab over a period of time

is the ans b.


if the satellite is at distant one earth radius form surface of earth calcualt eits velocity. radius of erath is 6400km. of i get this. but next part is the prob- given that the period of the moon is 30 days and it is 70 earth radii form centre of earth calcualte the period in days of the satellite in b.

 

[exa_boi87]

First Question:
As per Newtonian relativity, the two situations are virtually indistinguishable (travelling with constant velocity and stationary)

Second Question:
Ill do the whole thing to get all the info correct for calculation;

Radius of Earth: 6400km
Orbital Radius From Center: 12800km

Orbital Velocity = (sqrt) GM/r = (sqrt) (6.67x10^-11 x 5.983x10^24)/(12800 x 10^3)
Orbital Velocity = 3.12 x 10^7

now, they give you information regarding the moons motion, the only use we can make of this information is to calculate a 'question specific' mass of Earth (will end up being apox 6.0 x 10^24 anyway, but in such a question usually they want to see the process done).

Its sort of an odd question, what Id be doing is taking the period of the moon in relation to its orbital radius, and dividing it to get the smaller radius. Its messy I know, id be interested to see if anyone else comes up with an actual formula..

If you disregard the information they gave you re: the moon, you could get away with the 'orbital period' formula -

Orbital Velocity = 2(pie)r/T
3.12 x 10^7 = 2 x (pie) x (12800x10^3)/T
T = 0.39 hmmm

Keplers 3rd Law Of Periods =
r^3/T^2 = GM/4(pie)^2
(12800 x 10^3)/T^2 = (6.67 x 10^-11)x(5.983x10^24)/4(pie)^2
T^2 = 1.26 x 10^-6
T = 1.125 x 10^-3 (seconds)
T = 1.30 x 10^-8

then the other messy method (dividing radius')

2 earth radii / 70 earth radii = 1/35
therefore, orbital period 1/35th of Moons;
30/35 = 0.857 Days.

Sweet mother lol, if this post has done anything for your second question its probably provided some things to think about in the meantime before helper comes along

 

[KFunk]

You don't actually need to calculate that GM/4(pi)² stuff (it uses up a lot of time). What you're working with is the fact that r³/T² has a constant value for objects orbiting the earth hence:

r³/T² (satellite) = r³/T² (moon)

(2r)³/T² = (70r)³/30²

T² = (8x30²)/70³ = 0.02 days

Hence the period T of the satellite = 0.14 days


Which is quite different to the value in the last post. Damn .

 

[exa_boi87]

That makes alot more sense than my workings lol

 

[richz]

You don't actually need to calculate that GM/4(pi)² stuff (it uses up a lot of time). What you're working with is the fact that r³/T² has a constant value for objects orbiting the earth hence:

r³/T² (satellite) = r³/T² (moon)

(2r)³/T² = (70r)³/30²

T² = (8x30²)/70³ = 0.02 days

Hence the period T of the satellite = 0.14 days


Which is quite different to the value in the last post. Damn .

why is it 2r??

 

[KFunk]

why is it 2r??

'Cause the question says that the satellite is one earth radius from the 'surface' of the earth so I figured that it's two earth radii from the actual centre - hence 2r.

 

[exa_boi87]

Yeh its definately 2r, the equation takes 'r' as the distance from the objects center