Question

uniform circular motion & gravitation review sheet
name:
period: 4 date: 12/09/25
base your answers to questions 1-6 on the information below and on your knowledge of physics.
communication satellites and orbital mechanics
communication satellites enable global telecommunications, television broadcasting, and internet services. these satellites are often placed in geostationary orbits, where they remain above the same point on earths surface by matching earths rotation. understanding the physics of satellite orbits is essential for maintaining these critical communication networks.
a communication satellite orbits earth at an altitude of 3.59 × 10⁷ meters above earths surface. the satellite has a mass of 5.50 × 10³ kilograms and completes one orbit every 24 hours (8.64 × 10⁴ seconds). table 1 provides additional data about earth and the satellite system.
table 1: satellite and earth data

property	value
radius of earth	6.37 × 10⁶ m
satellite mass	5.50 × 10³ kg
orbital period	8.64 × 10⁴ s
distance from earths center	4.23 × 10⁷ m

1-2) develop a mathematical model to calculate the orbital speed of the communication satellite. show all work, including an equation and substitution with units. 2

Explanation:

Step1: Recall Orbital Speed Formula

Orbital speed \( v \) is the distance traveled in one orbit (circumference \( C = 2\pi r \)) divided by the orbital period \( T \). So the formula is \( v=\frac{2\pi r}{T} \), where \( r \) is the distance from Earth's center, and \( T \) is the orbital period.

Step2: Identify Values from Table

From Table 1: \( r = 4.23\times10^{7}\, \text{m} \), \( T = 8.64\times10^{4}\, \text{s} \).

Step3: Substitute Values into Formula

Substitute \( r \) and \( T \) into \( v=\frac{2\pi r}{T} \):
\( v=\frac{2\times\pi\times4.23\times10^{7}\, \text{m}}{8.64\times10^{4}\, \text{s}} \)
First, calculate the numerator: \( 2\times\pi\times4.23\times10^{7}\approx2.656\times10^{8}\, \text{m} \)
Then divide by the denominator: \( v\approx\frac{2.656\times10^{8}\, \text{m}}{8.64\times10^{4}\, \text{s}}\approx3074\, \text{m/s} \) (or simplify the exponents: \( \frac{2\times4.23\times10^{7}}{8.64\times10^{4}}\pi=\frac{8.46\times10^{3}}{8.64}\pi\approx0.979\times10^{3}\pi\approx3075\, \text{m/s} \), slight difference due to rounding).

Answer:

The orbital speed \( v \) is calculated using \( v = \frac{2\pi r}{T} \). Substituting \( r = 4.23\times10^{7}\, \text{m} \) and \( T = 8.64\times10^{4}\, \text{s} \), we get \( v\approx\frac{2\pi(4.23\times10^{7})}{8.64\times10^{4}}\approx3.07\times10^{3}\, \text{m/s} \) (or ~3070 m/s).