Question

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. these are called geosynchronous orbits. the radius of the earth is 6.37×106m, and the altitude of a geosynchronous orbit is 3.58×107m( ≈22000 miles)

Answer 1

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:

$\large {\boxed {a = \frac{ v^2 } { R } }$

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:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

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:

$F = ma$

$G \frac{ Mm}{(R+h)^2}=m v^2 \div (R+h)$

$G \frac{ M}{R+h} = v^2$

$v = \sqrt{ G \frac{M}{R+h}}$

$v = \sqrt{ 6.67 \times 10^{-11} \frac{5.98 \times 10^{24}}{6.37 \times 10^6 + 3.58 \times 10^7}}$

$\boxed{v = 3.08 \times 10^3 \texttt{ m/s}}$

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Next , we could calculate the acceleration of the satellite:

$a = v^2 \div ( R + h )$

$a = ( 3.08 \times 10^3 )^2 \div ( 6.37 \times 10^6 + 3.58 \times 10^7 )$

$\boxed{a \approx 0.225 \texttt{ m/s}^2}$

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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

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Answer 2

Answer: . 24.0 hr b. 0.223 m/s^2 c. 0 because there is no Normal Force on a satellite.