The satellite's orbital speed is about 3.08 × 10³ m/s
The acceleration of the satellite is about 0.225 m/s²
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Further explanation
Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
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Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
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Question:
a.What is the speed of a satellite in a geosynchronous orbit?
b.What is the magnitude of the acceleration of a satellite in a geosynchronous orbit?
Given:
height of the satellite = h = 22000 miles = 3.58 × 10⁷ m
mass of the earth = M = 5.98 × 10²⁴ kg
radius of the earth = R = 6.37 × 10⁶ m
Unknown:
Orbital Speed of the satellite = v = ?
Acceleration of the satellite = a = ?
Solution:
We will use this following formula to find the orbital speed:
[tex]F = ma[/tex]
[tex]G \frac{ Mm}{(R+h)^2}=m v^2 \div (R+h)[/tex]
[tex]G \frac{ M}{R+h} = v^2[/tex]
[tex]v = \sqrt{ G \frac{M}{R+h}}[/tex]
[tex]v = \sqrt{ 6.67 \times 10^{-11} \frac{5.98 \times 10^{24}}{6.37 \times 10^6 + 3.58 \times 10^7}}[/tex]
[tex]\boxed{v = 3.08 \times 10^3 \texttt{ m/s}}[/tex]
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Next , we could calculate the acceleration of the satellite:
[tex]a = v^2 \div ( R + h )[/tex]
[tex]a = ( 3.08 \times 10^3 )^2 \div ( 6.37 \times 10^6 + 3.58 \times 10^7 )[/tex]
[tex]\boxed{a \approx 0.225 \texttt{ m/s}^2}[/tex]
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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
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Answer details
Grade: High School
Subject: Physics
Chapter: Circular Motion
