The period of oscillation of a near-Earth satellite (neglecting atmospheric effects) is 84.3 min. What is the period of a near-Moon satellite? (R⊕= 6.37 × 10^6 m; RMoon= 1.74 × 10^6 m; m⊕ = 5.98 × 10^24 kg; mMoon = 7.36 × 10^22 kg.)
Recall that T^2 = 4π^2a^3/GM
That is, T^2M/a^3 = 4π^2/G, a constant
So, you want T such that
(7.36*10^22)T^2/(1.74*10^6)^3 = (5.98*10^24)(84.3)^2/(6.37*10^6)^3
Or, you know that
T^2 is proportional to M/a^3
so, the new (') values bear the relation to the old values:
(T'/T)^2 = (M'/M)(a'/a)^3
This kind of relation is useful if they say something like, what is the new period if you double the radius and triple the mass?
Well, you know, the near-Earth satellite is probably doing a little dance up there, enjoying the view. But when it comes to the Moon, things get a bit more interesting. So, let me put on my funny hat and give you the answer!
Now, the period of oscillation of a satellite depends on the gravitational field it's in. The gravitational field around the Moon is different than around Earth, you know? So we can use some physics magic to figure this out.
We know that the period of oscillation of a satellite around a planet or moon is given by:
T = 2π√(r³/GM)
Where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the planet or moon.
For the near-Earth satellite, we have:
T⊕ = 84.3 min = 84.3 × 60 = 5058 s
r⊕ = R⊕ + h, where R⊕ is the radius of Earth and h is the satellite's altitude above the surface. Neglecting atmospheric effects, let's assume h = 0 (on the surface).
Now, let's calculate the period of the near-Moon satellite:
rMoon = RMoon + h = RMoon + 0 = RMoon
MMoon = mMoon
Applying the formula, we get:
TMoon = 2π√(rMoon³ / GMoon)
Ah, but we need to find GMoon first!
GMoon = gMoon × RMoon², where gMoon is the gravitational field around the Moon.
To calculate gMoon, we can use Newton's law of universal gravitation:
gMoon = GMoon / RMoon² = G × mMoon / RMoon²
Finally, we can substitute GMoon and gMoon back into the equation for TMoon:
TMoon = 2π√(RMoon³ / (G × mMoon / RMoon²))
Now, you just have to plug in the values and crunch the numbers. I'll leave that part to you! Good luck! And remember, when you're dealing with the Moon, always watch out for werewolves... just in case.
To find the period of a near-Moon satellite, we will use Kepler's Third Law, which states that the square of the orbital period (T) of a satellite is directly proportional to the cube of the semi-major axis (a) of its orbit.
The semi-major axis can be calculated using the sum of the satellite's distance from the center of the Earth (R⊕) and the Moon's distance from the center of the Earth (RMoon).
a = R⊕ + RMoon
Next, we can calculate the period of the near-Moon satellite using the relationship provided by Kepler's Third Law:
T^2 = k*a^3
Where k is the constant of proportionality.
Since the period of the near-Earth satellite is given as 84.3 minutes, we can substitute the values into the equation and solve for k:
(84.3 min)^2 = k*(R⊕ + RMoon)^3
Simplifying the equation and solving for k gives:
k = (84.3 min)^2 / (R⊕ + RMoon)^3
Now, we can substitute the values for R⊕, RMoon, and k into the equation to find the period (T) of the near-Moon satellite:
T^2 = k*(R⊕ + RMoon)^3
T^2 = [(84.3 min)^2 / (6.37 × 10^6 m + 1.74 × 10^6 m)^3] * (6.37 × 10^6 m + 1.74 × 10^6 m)^3
Simplifying the equation gives:
T^2 = [(84.3 min)^2 / (8.11 × 10^6 m)^3] * (8.11 × 10^6 m)^3
T^2 = (84.3 min)^2
Taking the square root of both sides gives:
T = 84.3 min
Therefore, the period of a near-Moon satellite will also be 84.3 minutes.
To find the period of a near-Moon satellite, we can use Kepler's third law of planetary motion, which states that the square of the period of a planet (or satellite) is directly proportional to the cube of its average distance (semimajor axis) from the central body.
Let's denote the period of the near-Earth satellite as TE and the period of the near-Moon satellite as TM. We also need to find the semimajor axis of the near-Moon satellite, denoted as AM.
The average distance (semimajor axis) of the near-Earth satellite can be calculated using the equation:
AE = R⊕ + h
Where R⊕ is the radius of the Earth (6.37 × 10^6 m) and h is the height (altitude) of the satellite above the Earth's surface.
Since the near-Earth satellite is in a low Earth orbit, we can assume that h is much smaller than the radius of the Earth. Therefore, we can neglect h and consider AE ≈ R⊕.
Now, the average distance (semimajor axis) of the near-Moon satellite can be calculated as:
AM = RMoon + h
Similar to the near-Earth satellite, we can assume that h is much smaller than the radius of the Moon. Therefore, we can neglect h and consider AM ≈ RMoon.
Now, we can use the ratio of the average distances to find the ratio of the periods:
(TM/TE)^2 = (AM/AE)^3
Substituting the approximate values, we get:
(TM/TE)^2 = (RMoon/R⊕)^3
Calculating the right-hand side:
(TM/TE)^2 = (1.74 × 10^6 m / 6.37 × 10^6 m)^3
(TM/TE)^2 = (0.2725)^3
(TM/TE)^2 = 0.020085
Taking the square root of both sides:
TM/TE ≈ √0.020085
TM/TE ≈ 0.14182
Finally, we can find the period of the near-Moon satellite (TM) by multiplying the period of the near-Earth satellite (TE) by the ratio we just calculated:
TM = TE * (TM/TE)
Substituting the value of TE (84.3 min), we get:
TM = 84.3 min * 0.14182
TM ≈ 11.97 min
Therefore, the period of a near-Moon satellite is approximately 11.97 minutes.