A satellite is placed in orbit, with a radius that is half the radius of the Moon's orbit. Find the period of the satellite in units of the period of the Moon. (sketch and analyze the situation, show all working so I can understand how you arrived at your answer).
Ac = w^2 R
F = m a = m w^2 R = G m Me /R^2
so
w^2 = G Me/R^3
(2 pi /T)^2= G Me/R^3
(2 pi)^2/G Me = constant = T^2/R^3
so
Tmoon^2 / Rmoon^3 = Tsat^2/Rsat^3
Tsat^2 = Tmoon^2 (Rsat/Rmoon)^3
Tsat = Tmoon (1/2)^3/2 = 2sqrt 2 *Tmoon
Tsat = Tmoon (1/2)^3/2 = (1/2sqrt 2) *Tmoon
To find the period of the satellite in units of the period of the Moon, we need to understand the relationship between the period and the radius of an object in orbit.
Let's start by sketching the situation:
🌑
🌏 🛰️
Here, the 🌑 represents the Moon, the 🌏 represents the Earth, and the 🛰️ represents the satellite.
We are given that the radius of the satellite's orbit is half the radius of the Moon's orbit. Let's say the radius of the Moon's orbit is r, thus the radius of the satellite's orbit is 0.5r.
Now, let's analyze the situation using the concept of centripetal force:
The centripetal force required to keep an object in circular motion is given by the equation F = (m*v^2) / r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the orbit.
We know that the gravitational force between two objects is given by F = (G * m * M) / r^2, where G is the gravitational constant, m is the mass of the orbiting object, M is the mass of the object being orbited, and r is the distance between the center of both objects.
Setting these two equations equal to each other, we have:
(G * m * M) / r^2 = (m * v^2) / r
The mass of the satellite cancels out, leaving:
(G * M) / r = v^2
Since the speed of an object in circular motion is given by v = (2 * π * r) / T, where T is the period of the orbit, we can substitute this into the equation above:
(G * M) / r = [(2 * π * r) / T]^2
Simplifying the equation, we get:
(G * M) / r = 4 * π^2 * r^2 / T^2
Cross multiplying, we have:
(G * M * T^2) / (4 * π^2 * r) = r^2
Taking the square root of both sides, we get:
T / (2 * π) = √(r / (G * M))
Solving for T, we have:
T = 2 * π * √(r / (G * M))
Now, since we want to find the period of the satellite relative to the period of the Moon, we can express the radius of the satellite's orbit in terms of the radius of the Moon's orbit (r_moon):
r = 0.5 * r_moon
Substituting this back into the equation for T:
T = 2 * π * √((0.5 * r_moon) / (G * M))
Simplifying further:
T = π * √(0.5 * r_moon / (G * M))
Now, the period of the Moon is approximately 27.3 days, so we can express it as:
T_moon = 27.3 days
Now, we can find the period of the satellite in units of the period of the Moon:
T / T_moon = (π * √(0.5 * r_moon / (G * M))) / T_moon
However, since we are not given specific values for the radius of the Moon's orbit or the masses of the Moon and Earth, we cannot compute the exact value of T / T_moon. We can only express it in terms of these unknown values.
Therefore, the final answer is:
T / T_moon = (π * √(0.5 * r_moon / (G * M))) / T_moon
To find the period of the satellite in units of the Moon's period, let's first understand the situation.
We have a satellite placed in orbit with a radius that is half the radius of the Moon's orbit. Let's assume the satellite and the Moon orbit around a central body (like Earth) in a circular path.
To analyze the situation, let's consider the following:
1. Radius of the Moon's orbit: let's denote it as Rm.
2. Radius of the satellite's orbit: let's denote it as Rs. Given that the satellite's radius is half the Moon's radius, Rs = (1/2) * Rm.
3. Period of the Moon's orbit: let's denote it as Tm.
4. Period of the satellite's orbit: let's denote it as Ts (what we are trying to find).
Now, we can use the concept of Kepler's Third Law, also known as the Law of Harmonies, which states that the square of the period of an orbit is proportional to the cube of its radius.
Mathematically, this can be represented as:
(T1 / T2)^2 = (R1 / R2)^3
Where T1 and T2 are the periods of two different orbits, and R1 and R2 are the radii of those orbits.
In our case, we can use this equation to relate the period of the satellite's orbit (Ts) to the period of the Moon's orbit (Tm) and their respective radii (Rs and Rm):
(Ts / Tm)^2 = (Rs / Rm)^3
Substituting the values:
[(Ts / Tm)^2] = [((1/2) * Rm) / Rm]^3
= (1/2)^3
= 1/8
To solve for Ts / Tm, we can take the square root of both sides:
Ts / Tm = √(1/8)
Ts = (√(1/8)) * Tm
The period of the satellite (Ts) in units of the Moon's period (Tm) is (√(1/8)) times Tm.
Simplifying further, we can evaluate (√(1/8)):
√(1/8) = √(1) / √(8)
= 1 / 2√(2)
= 1 / (2 * √(2))
= 1 / (2 * √(2)) * (√(2) / √(2))
= √(2) / (2 * √(2) * √(2))
= √(2) / (2 * 2)
= √(2) / 4
Therefore, the period of the satellite (Ts) in units of the Moon's period (Tm) is (√(2) / 4) times Tm.
In summary, Ts = (√(2) / 4) * Tm.