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USPshnik [31]
3 years ago
11

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

2 km/s. The minimum speed needed to escape from the surface of the planet is 14.9 km/s, and G = 6.67 × 10-11 N · m2/kg2. The orbital period of the satellite is closest to
Physics
1 answer:
Ksju [112]3 years ago
3 0

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

v =  \sqrt{\frac{2GM}{R}}

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

v = 14.9km/h = 14900m/s

Through this equation we can find the mass of the Planet in function of the distance, therefore

M = \frac{v^2R}{2G}

M = \frac{14900^2R}{2(6.67*10^{-11})}

M = 16.64*10^{17}R

The orbital velocity is

v_o = \sqrt{\frac{GM}{R+h}}

9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}

11.1*10^7R = (R+15000*10^3)(9200)^2

2.64*10^7R = 12.69*10^{13}

R = 4.81*10^6m

The time period of revolution is,

T = \frac{2\pi(R+h)}{v_o}

T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}

T = 4307s

T = 72min = 1hour12min

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

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