Question

Calculate the period of a satellite orbiting the moon, 94 km above the moon's surface. ignore effects of the earth. the radius of the moon is 1740 km.

Answer 1

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

Answer 2

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:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

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:

$F = G \frac{m_{moon} \times m_{satellite}}{R^2}$

$m_{satellite} \times \omega^2 \times R = G \frac{m_{moon} \times m_{satellite}}{R^2}$

$\omega^2 \times R = G \frac{m_{moon}}{R^2}$

$\omega^2 = G \frac{m_{moon}}{R^3}$

$\omega = \sqrt{G \frac{m_{moon}}{R^3}}$

$\omega = \sqrt{6.67 \times 10^{-11} \frac{7.35 \times 10^{22}}{(1.834 \times 10^6)^3}}$

$\omega = 8.91 \times 10^{-4}$

$\frac{2 \pi}{T} = 8.91 \times 10^{-4}$

$T = \frac{2 \pi}{8.91 \times 10^{-4}}$

$\boxed {T \approx 7048 ~ seconds}$

Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : 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