mikynya9510
contestada

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Determine the orbital period of the satellite.

Respuesta :

 g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)
nuhulawal20

The orbital period of the artificial satellite is 1.5 hours.

Given the data in the question;

Acceleration due to gravity; [tex]a = 9.00 m/s^2[/tex]

Orbital period of the satellite; [tex]T =\ ?[/tex]

To determine the orbital period of the satellite, we make use of Kepler's third law:

[tex]T^2= [\frac{4\pi ^2}{GM_E}]r^3[/tex]

First we find the orbital radius of the artificial satellite

Using Newton's law of universal gravitation

[tex]F = \frac{Gm_Em}{r^2}[/tex]

Where m is mass of satellite, [tex]m_E[/tex]  is mass of earth ( [tex]5.972 * 10^{24} kg[/tex] ), G is the universal gravitational constant( [tex]6.67430*10^{-11} N.m^2/kg^2[/tex] ) and r is the orbital radius.

We know that, Force; [tex]F = mass *gravity = mg[/tex]

So

[tex]mg = \frac{Gm_Em}{r^2}\\\\g = \frac{Gm_E}{r^2}\\\\r^2 = \frac{Gm_E}{g}\\\\r = \sqrt{\frac{Gm_E}{g}}[/tex]

We substitute in the values

[tex]r = \sqrt{\frac{(6.67430*10^{-11}N.m^2/kg^2)(5.972*10^{24}kg}{9.00m/s^2}}\\\\r = 6.65*10^6 m[/tex]

Now,  using the of Kepler's third law:

[tex]T^2= [\frac{4\pi ^2}{GM_E}]r^3[/tex]

We substitute our values into the equation

[tex]T^2 = [\frac{4\pi ^2}{6.67430*10^{-11}N.m^2/kg^2)(5.972*10^{24}kg} ](6.65*10^6m)^3\\\\T^2 = 2.9127*10^{7}s^2\\\\T = \sqrt{2.9127*10^{7}s^2}\\\\T = 5396.94s\\\\T = 1.5hrs[/tex]

Therefore, the orbital period of the artificial satellite is 1.5 hours.

Learn more: https://brainly.com/question/21049176

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