A satellite in orbit around the earth has a period of one hour. An identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. What is the
period of the second satellite?
since the period is proportional to r^(3/2), replacing r with (9r) produces a period 9^(3/2) = 27 times longer.
To determine the period of the second satellite, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution of a celestial body is proportional to the cube of its average distance from the celestial body it is orbiting.
In this case, we have two satellites: the first one with a period of one hour and an orbit radius, and the second one with a larger orbit radius that is nine times larger than the first satellite.
Let's denote the period of the second satellite as T2 and the orbit radius of the first satellite as r1. Since the orbit radius of the second satellite is nine times larger, we can say that the orbit radius of the second satellite is 9*r1.
According to Kepler's Third Law, we have:
(T1^2) / (r1^3) = (T2^2) / (r2^3)
Substituting the given values:
(1^2) / (r1^3) = (T2^2) / ((9*r1)^3)
Simplifying further:
1 / (r1^3) = (T2^2) / (729*r1^3)
Cross-multiplying and canceling out r1^3:
r1^3 = 729*T2^2
Now, let's solve this equation to find the value of T2:
T2^2 = r1^3 / 729
Taking the square root of both sides:
T2 = sqrt(r1^3 / 729)
Since we know that the period of the first satellite, T1, is one hour, we can substitute it into the equation:
T2 = sqrt(1^3 / 729)
T2 = sqrt(1 / 729)
T2 = 1 / 27
Therefore, the period of the second satellite is 1/27 hour or approximately 2.222 minutes.
The period of an object in orbit around a celestial body is determined by its distance from the center of the celestial body.
The period (T) of an object in circular orbit can be found using Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit:
T² ∝ r³
Where T is the period and r is the radius of the orbit.
For the first satellite, let's assume its radius is 'r', and its period is 1 hour. According to the problem, the second satellite has a radius that is nine times larger than the first satellite. So, the radius of the second satellite is 9r.
Using Kepler's third law, we can write the equation for the first satellite as:
(1)² ∝ r³
Simplifying, we get:
1 ∝ r³
Now, let's calculate the period of the second satellite:
T² ∝ (9r)³
T² ∝ 729r³
Since the two satellites are identical, the proportionality factor remains the same. Thus, we have:
1 ∝ 729r³
T² ∝ 729r³
Comparing the two equations, we can see that the period of the second satellite, T², will be 729 times larger than the period of the first satellite, 1.
Therefore, the period of the second satellite would be the square root of 729, which is 27 hours.
So, the period of the second satellite is 27 hours.