Question

An earth satellite moves in a circular orbit at a speed of 5000 m/s. part a what is its orbital period? express your answer using two significant figures

Answer 1

Its orbital period is about 5.6 hours

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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 !

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Given:

speed of satellite = v = 5000 m/s

mass of earth = M = 6 × 10²⁴ kg

Asked:

orbital period = T = ?

Solution:

$\Sigma F = ma$

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

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

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

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

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

$\frac{ 2 \pi R }{ v } = (2 \pi G \frac{ M } { v^2 } ) \div v$

$T = 2 \pi G \frac{ M } { v^3 }$

$T = 2 \pi \times 6.67 \times 10^{-11} \times \frac{ 6 \times 10^{24} } { 5000^3 }$

$T \approx 20116 \texttt{ s}$

$T \approx 5.6 \texttt{ hours}$

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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: Gravitational Fields

Answer 2

r = GM/v^2 r is the distance of the satellite from the center of Earth in meters G is newtons constant, 6.67428 x 10^-11 M is the mass of Earth, 5.9742 x 10^24 v is the orbital speed, which we know is 5,000 m/s r = 6.67428 x 10^-11 x 5.9742 x 10^24/5,000^2 r = 8,137,445 meters from the center of the Earth and : 8,137,445 - 6,378,100 = 1,749,345 meters from the surface of the Earth Now that we know the distance we can calculate the orbital period. Time = Distance/speed distance is the circumference of the orbit, 2Ď€ x 8,137,445 = 51,129,074 meters Time = (51,129,074)/5,000 Time = 10,225 seconds = 2.84 hours so, orbital period is 2.84 hrs