contestada

A particular satellite was placed in a circular orbit about 529 mi above Earth.
(a) Determine the orbital speed of the satellite. (m/s)
(b) Determine the time required for one complete revolution. (sec)

Respuesta :

sebasacevedop

Explanation:

(a) The orbital speed is defined as:

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

Here, G is the gravitational constant, M is the Earth's mass and r is the radius of the circular orbit

[tex]529mi*\frac{1609.34m}{1mi}=8.51*10^5m[/tex]

[tex]v=\sqrt{\frac{(6.67*10^{-11}\frac{N\cdot m^2}{kg})(5.97*10^{24}kg)}{8.51*10^5m}}\\v=2.16\frac{m}{s}[/tex]

(b) The time required for one complete revolution is the period of the circular motion and is given by:

[tex]T=\frac{2\pi r}{v}\\T=\frac{2\pi(8.51*10^5m)}{2.51*10^4\frac{m}{s}}\\T=247.55s[/tex]

hamzaahmeds

This question involves the concepts of orbital speed and time period.

(a) The orbital speed of the satellite is "7666 m/s".

(b) The time period of the satellite is "5925.83 s".

(a)

The orbital speed is given by the following formula:

[tex]v-\sqrt{\frac{GM}{R}}\\\\[/tex]

where,

v = orbital speed = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = mass of earth = 5.97 x 10²⁴ kg

R = distance of satellite from center of earth

R = radius of earth + (529 mi)[tex]\frac{1609.34\ m}{1\ mi}[/tex] = 6.38 x 10⁶ m + 0.85 x 10⁶ m

R = 7.23 x 10⁶ m

Therefore,

[tex]v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2\kg^2)(6.37\ x\ 10^{24}\ kg)}{7.23\ x\ 10^6\ m}}\\\\[/tex]

v = 7666 m/s

(b)

The time period of the satellite can be given by the following formula:

[tex]T=\frac{2\pi R}{v}\\\\T=\frac{2\pi (7.23\ x\ 10^6\ m)}{7666\ m/s}[/tex]

T = 5925.83 s = 98.76 min = 1.65 h

Learn more about orbital speed here:

https://brainly.com/question/541239?referrer=searchResults

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