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A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbiting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.

6,800 m/s
7,200 m/s
7,500 m/s
7,900 m/s

Respuesta :

MathPhys

Answer:

7500 m/s

Explanation:

Centripetal acceleration = gravity

v² / r = GM / r²

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m

v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))

v = 7500

The orbital velocity is 7500 m/s.

The orbital velocity is 7500 m/s.

What is orbital velocity ?

Orbital velocity is defined as the minimum velocity a body must maintain to stay in orbit.

Given

h = 6.8 E 5 m = 6.8 * [tex]10^{5}[/tex] m

mass of satellite = 5.6 * [tex]10^{5}[/tex] kg

Earth's mass : m(e)  = 5.98 E 24 kg

Earth's radius ; R(e) = 6.357 E 6 m

orbital velocity :v(orbiting ) = [tex]\sqrt{Gm(e) / R(e) + h}[/tex]

R(e) + h  = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m

v = √(6.67×10⁻¹¹ * (5.98×10²⁴)) / (7.037×10⁶))

v = 7500 m/s

The orbital velocity is 7500 m/s.

learn more about orbital velocity

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