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.
a) 6,800 m/s
b) 7,200 m/s
c) 7,500 m/s
d) 7,900 m/s

C) 7,500 m/s

Explanation:

The satellite's acceleration due to gravity equals its centripetal acceleration.

v² / r = GM / r²

Solving for velocity:

v² = GM / r

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.357×10⁶ m + 6.8×10⁵ m = 7.037×10⁶ m

Substituting the values:

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

v = √(5.67×10⁷)

v = 7500 m/s

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..

i think it's 35 meters