Calculating earth's speed using radius and speed of light, etc.

AI Thread Summary
Ole Roemer's observation of a 14-second delay in Io's disappearance provides a basis for calculating light travel distance, which is approximately 4.2 billion meters. This distance, when divided by Io's orbital period of 42.5 hours, yields an Earth speed of 27.45 km/s. The discussion highlights the importance of using the circumference of Earth's orbit, calculated as 2π times the orbital radius, to find the correct speed. Participants clarify that light travels across the diameter rather than the circumference, emphasizing the need for angular velocity in this context. Overall, the calculations and discussions focus on accurately determining Earth's speed in relation to its orbital mechanics.
DDRchick
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Ole Roemer found that the average increased delay in the disappearance of Io from one orbit around Jupiter to the next is 14 s.
(a) How far does light travel in 14 s?
1 m

(b) Each orbit of Io takes 42.5 h. Earth travels the distance calculated in part (a) in 42.5 h. Find the speed of Earth in km/s.
2 km/s

(c) Check to make sure that your answer for part (b) is reasonable. Calculate Earth's speed in orbit using the orbital radius, 1.5 108 km, and the period, one year.




d=vt
speed of light = 3x10^8




For part (a) i got 4.2e9, which was correct. d=(3x10^8)(14)
For part (b) I got 27.45 km/s which was correct. 4.2e9m = 4.2e6km
Part (c) I don't know...
 
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The Earth goes around the sun in a circle. Find the distance around the circle. The time taken to go around, in seconds. Use v = d/t to find the speed.
 
Well I figure, radius is half a diameter, right?
So i multiplied the radius (1.5 x 10^8) by two.
then i found the number of seconds in a year, which is 31,536,000.
I even tried just doing (1.5x10^8)/# seconds in a year
and it wa s still wrong.
:(


**Because it won't let me edit the first post...**
orbital radius, 1.5 108 km = 1.5x10^8
 
Last edited:
DDRchick said:
Well I figure, radius is half a diameter, right?
So i multiplied the radius (1.5 x 10^8) by two.
then i found the number of seconds in a year, which is 31,536,000.

Isn't the circumference of a circle = 2πr ?
 
Light doesn't travel around the circumference of a circle- travels across the diameter.
 
I used LowlyPion's equation and plugged in the radius, and then divided by the number of seconds.
It marked it correct. :D
Thanks so much!
 
HallsofIvy said:
Light doesn't travel around the circumference of a circle- travels across the diameter.

That may well be, but the question is asking for Earth's speed, and the circumference/period. Earth's speed is relevant for explaining the 14 s interval, and represents an exceedingly small arc of Earth's orbit right?
 
hey it is the angular velocity you need to consider i.e omega,not linear velocity
then v=d/t cannot be used .
 
Just curious would you have the θ for 14 sec divided by a year load of seconds?

And could you identify the error difference between the arc of θ, and the chord of θ?
 
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