Asatellite 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

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.

less than one you can't create energy

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft