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A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

A) Determine the radius of its circular orbit.(meters)

B) Determine the magnitude of the Earth's gravitational force on the satellite. (N)

C) Determine the altitude of the satellite. (m)
Expert Answer

Respuesta :

Answer:

(a)  Radius of orbit will be [tex]=7.32\times10^6m[/tex]

(b) Earth gravitational force will be [tex]=4.18\times 10^4N[/tex]

(C) Height will be [tex]0.92\times 10^6m[/tex]

Explanation:

We have given

Mass of the earth, [tex]M=6\times 10^{24}kg[/tex]

Mass of the satellite, m = 5600 kg

Radius of earth, [tex]R=6.4\times 10^6m[/tex]

Time period T = 6200 sec

We know that [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec[/tex]

Now

(a) We know that [tex]\omega ^2=\frac{GM}{R^3}[/tex]

[tex]R^3=\frac{GM}{\omega ^2}[/tex]  

[tex]R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}[/tex]

[tex]R^3=3.92\times 10^{20}[/tex]

Radius of the orbit [tex]R=7.32\times 10^6m[/tex]

(b)

Force [tex]F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N[/tex]

(c)

Altitude [tex]h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m[/tex]

onyebuchinnaji

(a) The radius of its circular orbit is 7.308 x 10⁶ m.

(b) The magnitude of the Earth's gravitational force on the satellite is 41,753.3 N.

(c) The altitude of the satellite is 9.08 x 10⁵ m.

The given parameters;

  • mass of the satellite, m = 5600 kg
  • period of the satellite, T = 6200 s
  • mass of earth, m = 5.97 x 10²⁴ kg

The angular speed of the satellite is calculated as follows;

[tex]\omega = \frac{2\pi }{T} \\\\\omega = \frac{2\pi }{6200} = 0.00101 \ rad/s[/tex]

The radius of the circular orbit is calculated as follows;

[tex]m \omega ^2 R = \frac{GMm}{R^2} \\\\\omega ^2 = \frac{GM}{R^3}\\\\R^3 = \frac{GM}{\omega ^2}\\\\R = \sqrt[3]{\frac{GM}{\omega ^2}} \\\\R = \sqrt[3]{\frac{(6.67\times 10^{-11}) \times (5.97 \times 10^{24}))}{(0.00101) ^2}}\\\\R = 7.308 \times 10^6 \ m[/tex]

The magnitude of the Earth's gravitational force on the satellite is calculated as follows;

[tex]F = \frac{GMm}{R^2} \\\\F = \frac{(6.67\times 10^{-11}) \times (5.97 \times 10^{24}) \times (5600)}{(7.308\times 10^6)^2} \\\\F = 41,753.3 \ N[/tex]

The altitude of the satellite is calculated as follows;

[tex]h = (7.308 \times 10^6) - (radius \ of \ earth)\\\\h = 7.308 \times 10^6 - 6.4\times 10^6\\\\h = 9.08 \times 10^5 \ m[/tex]

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