Respuesta :
Answer:
W = 24.28 kN
Explanation:
given,
Mass of satellite = 5850 Kg
height , h = 4.1 x 10⁵ m
Radius of planet = 4.15 x 10⁶ m
Time period = 2 h
= 2 x 3600 = 7200 s
Time period of satellite
[tex]T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
R is the radius of planet
h is the height of satellite
[tex]T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}[/tex]
now calculation of acceleration due to gravity
[tex]g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}[/tex]
[tex]g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}[/tex]
g = 4.15 m/s²
True weight of satellite
W = m g
W = 5850 x 4.15
W = 24277.5 N
W = 24.28 kN
True weight of the satellite is W = 24.28 kN
The true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
What is the time period of satellites?
Time period of satellites is the total time taken by a satellite to complete a full orbit around a body. It can be given as,
[tex]T=\dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
Here, (R) is the radius of the body, and (g) is the gravitational acceleration force.
In a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is 2 hours and the radius of the planet is 4.15 x 10 to the 6th power m.
To find the weight of the satellite, first find the value of gravitation acceleration using the time period formula as,
[tex]2=\dfrac{4\pi^2}{4.15\times10^6}\sqrt{\dfrac{(4.15\times10^6+4.1\times10^5)^3}{g}}\\g=4.15\rm m/s^2[/tex]
The weight of the body is mass time gravity. As the satellite has a mass of 5850 kg and value of g is 4.15 m/s². Thus, the weight of it is,
[tex]W=5850\times4.15\\W=24277.5\rm N\\W=24.28\rm \; kN[/tex]
Thus, the true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
Learn more about the time period of satellites here
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