Get comprehensive answers to your questions with the help of IDNLearn.com's community. Our experts are ready to provide prompt and detailed answers to any questions you may have.
The gravitational force exerted by the moon on the satellite is such that
F = G M m / R ² = m a → a = G M / R ²
where a is the satellite's centripetal acceleration, given by
a = v ² / R
The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that
v = 2πR / T → a = 4π ² R / T ²
Substitute this into the first equation and solve for T :
4π ² R / T ² = G M / R ²
4π ² R ³ = G M T ²
T ² = 4π ² R ³ / (G M )
T = √(4π ² R ³ / (G M ))
T = 2πR √(R / (G M ))
The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:
[tex] F = \frac{GmM}{R^{2}} [/tex] (1)
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
R: is the distance between the satellite and the center of the moon
m: is the satellite's mass
M: is the moon's mass
The gravitational force is also equal to the centripetal force:
[tex] F = ma_{c} [/tex] (2)
The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):
[tex] a_{c} = \frac{v^{2}}{R} [/tex] (3)
And from the tangential velocity we can find the period:
[tex] v = \omega R = \frac{2\pi R}{T} [/tex] (4)
Where:
ω: is the angular speed = 2π/T
By entering equations (4) and (3) into (2), we have:
[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex] (5)
By equating (5) and (1), we get:
[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]
[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]
[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]
[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]
Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
Find more here:
I hope it helps you!
