i have a few questions ranged over all 3 modules.. ple help.. (1 Viewer)

[kooltrainer]

1)First.. geostationary satallite has an altitude of 36000Km and has a period of 24 hours yes?.. so when i used this formula [r ^3]/[T^2] = GM/[4pi^2] ... i subbed r = 36000000m .. and got T = 67841.4 s .. which isnt 24 hours.. so whys that??

2) A geostationary satellite is placed at a distance of about 36000km from centre of earth. Calculate the period (in days) of a satellite placed at 72000 km from centre of earth.

3) An electron follows a circular path in a uniform magnetic field of strength 2.7x10^-2 T. The speed of the electron is 5.3x10^7 m/s.After 2 complete orbits, how much energy does the electron gain?

a)0J
b)1.28x10^-15 J
c)1.79x10^-21 J
d) 1.43 x 10^6 J


4) A beam of monochromatic light falls onto a cold, perfect black body and imparts 0.1 mW of power to it. The wavelength of the light is 5x10^-7m. Calculate the number of photons per second striking the black body.

 

[Pwnage101]

1) altitude DOES NOT equal radius - u MUST ADD THE RADIUS OF THE EARTH TO THE ALTITUDE TO GET A VALUE FOR 'r' TO SUB IN

radius of earth approx = 6 400 000m
altitude =36 000 000m
therefore r = 42 400 000m
G=6.672*10^-11
M= 6.0*10^24kg

invert formula:
[T^2]/[r ^3] = [4pi^2]/GM
[T^2] = [4pi^2]*[r ^3]/GM
[T^2] = [sub in values] = 7,519,348,649
T = 86,714s
= 1445 mins
= 24.08hours
=1 day

2). teh Q is wrong, it is 36 000km above the GROUND, not the centre of the earth, but IF TEH Q WERE CORRECT u would use keplers laws - r^3/T^2 is constant for orbiting bodies around an object

 

[kooltrainer]

o right.. that made it a bit more clear... what about question 3 , 4?

 

[xiao1985]

Q3, I'm sure this question is a tricky one, since when a charged particle accelerates, it gives off EMR, hence it really should lose energy. But ignoring that, it shouldn't gain or lose any energy, since it's undergoing circular motion, and the tangential velocity won't change.

Q4
P = energy per second
E= h f

c = f lambda
f = c/ lambda

therefore E = h c/ lambda = energy of a single photon
= 6.626 e-34 x 3 e8 / 5e-7 = 3.9756 e-19

number of photon/sec = (energy/sec) / (energy/photon)
=0.1e-3 / 3.9756e-19 = 2.515 e14 photons per second

 

[kooltrainer]

Q3, I'm sure this question is a tricky one, since when a charged particle accelerates, it gives off EMR, hence it really should lose energy. But ignoring that, it shouldn't gain or lose any energy, since it's undergoing circular motion, and the tangential velocity won't change.

so when would it lose / gain energy? when tangential velocity changes?? how? i dun understand...

 

[xiao1985]

First year uni phys stuff, almost forgetting. But when a charged particle accelerates, it emit EMR. ie, energy being emitted/lost by the electron. Due to the conservation of energy, itself must lose it's kinetic energy.

This could be observed by the loss of the kinetic energy.

But I should stress that it's not something which I remember very vividly, so I could be wrong.

 

[kooltrainer]

oh .. what course in uni learns physics stuff?? and which one is more likely to get a job afterwords..
i find that courses mostly involve chemistry rather than physics..

 

[cheney31]

oh .. what course in uni learns physics stuff?? and which one is more likely to get a job afterwords..
i find that courses mostly involve chemistry rather than physics..

as far as i know only engineering and physics itself actually need phy.. but correctly me if im wrong..

 

[xiao1985]

as far as i know only engineering and physics itself actually need phy.. but correctly me if im wrong..

astronomy, software design, project management, estimation, projection, modelling, warfare (arteliary), aeroneutical (wait, that's engineering)