Wavelength Speed Question

  • Thread starter Thread starter jemjabella42
  • Start date Start date
  • Tags Tags
    Speed Wavelength
AI Thread Summary
Radio waves travel at the speed of light, and a signal sent from the ground to a satellite in geosynchronous orbit (36,000 km above Earth) must account for the round trip distance. The equation used is speed equals distance over time, leading to the calculation of time taken for the signal to travel to the satellite and back. The conversion from kilometers to meters is crucial, as 36,000 km equals 36 million meters. The discussion highlights the confusion around isolating variables and dimensional analysis, emphasizing the importance of understanding unit conversions in physics calculations.
jemjabella42
Messages
11
Reaction score
0
Radio waves travel at the speed of light. A satellite is in a "geosynchronous orbit." A radio signal is sent from the ground to the satellite and then the satellite sends the signal back down to the ground. Satellites in geosynchronous orbit are 36,000km above the surface of the earth. How much time does it take for a signal to go from the ground to the satellite and back to the ground?

I plugged numbers into the equation speed=distance/time to get:
2.99792458x10^8 meters per second=36,000km/T

Then, I took both sides multiplied by T to get:
2.99792458x10^8 meters/sec(T)=36,000km

I am not sure how to isolate my variable from here because I'm letting the labels throw me off. I know 1km=1000meters. Do I need to use dimensional analysis to get past this step? It is very confusing for me and I don't know why! It seems like it should be simple math but my brain is just not used to it.
 
Physics news on Phys.org
Here we go ... like peeling layers off an onion.

v = d/t, where we use 't' for time and reserving 'T' for special intervals of time like the period of a pendulum.

then
3 x 10^8 = 2 x 36 x10^6/t thus converting km into m, noting that the signal has to return back.

Now let us 'peel off' t from the RHS

3 x 10^8 x t = 2 x 36 x10^6 .....multiplying both sides by t

then we 'peel off 3 x 10^8 from the LHS

t = 2 x 36 x10^6/3 x 10^8.....dividing both sides by 3 x 10^8

t = ...
 
grzz said:
Here we go ... like peeling layers off an onion.

v = d/t, where we use 't' for time and reserving 'T' for special intervals of time like the period of a pendulum.

then
3 x 10^8 = 2 x 36 x10^6/t thus converting km into m, noting that the signal has to return back.

Now let us 'peel off' t from the RHS

3 x 10^8 x t = 2 x 36 x10^6 .....multiplying both sides by t

then we 'peel off 3 x 10^8 from the LHS

t = 2 x 36 x10^6/3 x 10^8.....dividing both sides by 3 x 10^8

t = ...

I get confused when converting. I understand now why I need to multiply by 2 but how exactly do I arrive at 36x10^6? Could you explain this part in more detail please?
 
What do you mean by "arrive at 36x10^6"? It is given in the problem that the satellite is 36000 km above the earth. Presumably you know that there are 1000 meters in a kilometer (if you don't then it is worth memorizing "kilo= 1000") so it is 36000*(1000)= 36000000= 36*10000000= 36*10^6 meters above the earth.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top