Respuesta :
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
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:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
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:
[tex]\Sigma F = ma[/tex]
[tex]G \frac{ M m} { R^2 } = m \frac{v^2}{R}[/tex]
[tex]G \frac{ M } { R^2 } = \frac{v^2}{R}[/tex]
[tex]G \frac{ M } { R } = v^2 [/tex]
[tex]R = G \frac{ M } { v^2 } [/tex]
[tex]2 \pi R = 2 \pi G \frac{ M } { v^2 } [/tex]
[tex]\frac{ 2 \pi R }{ v } = (2 \pi G \frac{ M } { v^2 } ) \div v[/tex]
[tex]T = 2 \pi G \frac{ M } { v^3 }[/tex]
[tex]T = 2 \pi \times 6.67 \times 10^{-11} \times \frac{ 6 \times 10^{24} } { 5000^3 }[/tex]
[tex]T \approx 20116 \texttt{ s}[/tex]
[tex]T \approx 5.6 \texttt{ hours}[/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: Gravitational Fields
