Two satellites, A and B, are in different circular orbits about the earth. The orbital speed of satellite A is eighty-two times that of satellite B. Find the ratio (TA/TB) of the periods of the satellites.
To find the ratio of the periods of the satellites, we first need to understand the relationship between the orbital speed and the period of a satellite in a circular orbit.
The orbital speed of a satellite is the speed at which it moves in its orbit, while the period is the time it takes to complete one orbit.
The orbital speed of a satellite is directly related to the radius of its orbit, while the period is inversely related to the orbital speed. This means that as the orbital speed increases, the period decreases, and vice versa.
Let's represent the orbital speed of satellite A as VA and the orbital speed of satellite B as VB.
Given that VA = 82 * VB, we can set up a ratio:
VA / VB = 82 / 1
Using the relationship between orbital speed and period, we know that:
VA = 2πRA / TA
VB = 2πRB / TB
Where RA and RB are the radii of the orbits of satellites A and B, respectively, and TA and TB are their respective periods.
Since the satellites are in circular orbit around the Earth, we can assume that their radii are constant and equal to the radius of the Earth, which we can represent as R.
RA = RB = R
Substituting the values into our equations, we get:
2πR / TA = 82 * (2πR / TB)
Simplifying the equation, we can cancel out the 2πR terms:
1 / TA = 82 / TB
Now, we have:
TA / TB = 1 / (82 / TB)
To further simplify, we can invert the fraction on the right side:
TA / TB = TB / 82
Therefore, the ratio of the periods of the satellites (TA / TB) is TB / 82.
This means that the period of satellite A is TB times 82.