Respuesta :
First we calculate the effective radius R:
R = 1740+94 = 1834 km = 1.834E6 m
Then taking the mass of
the moon:
M = mass of moon = 7.348E22 kg
We then calculate the
period using the formula:
T = 2π√[R³/(GM)] = 7049.3 sec = 1.96 hr
The period of a satellite orbiting the moon is 7048 s
Further explanation
Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
To find the period of the satellite can be carried out in the following way:
[tex]F = G \frac{m_{moon} \times m_{satellite}}{R^2}[/tex]
[tex]m_{satellite} \times \omega^2 \times R = G \frac{m_{moon} \times m_{satellite}}{R^2}[/tex]
[tex]\omega^2 \times R = G \frac{m_{moon}}{R^2}[/tex]
[tex]\omega^2 = G \frac{m_{moon}}{R^3}[/tex]
[tex]\omega = \sqrt{G \frac{m_{moon}}{R^3}}[/tex]
[tex]\omega = \sqrt{6.67 \times 10^{-11} \frac{7.35 \times 10^{22}}{(1.834 \times 10^6)^3}}[/tex]
[tex]\omega = 8.91 \times 10^{-4}[/tex]
[tex]\frac{2 \pi}{T} = 8.91 \times 10^{-4}[/tex]
[tex]T = \frac{2 \pi}{8.91 \times 10^{-4}}[/tex]
[tex]\boxed {T \approx 7048 ~ seconds}[/tex]
Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
Answer details
Grade: High School
Subject: Physics
Chapter: Gravitational Fields
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
