Consider a satelite in a low altitude orbit around the Earth. The gravitational acceleration felt by the satelite is very close to the gravitational acceleration felt by objects on the surface of the earth: 9.8m/s^2 and is directed toward the center of the Earth. The orbit of the satelite is a perfect circle centered on the center of the Earth. There are no forces other than gravity acting on the satelite. Find how fast the satelite must be moving in order to remain in orbit. Explain all steps you take in order to carry out this calculation using concepts we have defined in this class.

The orbital speed of the satellite around the earth in other to remain in perfect circular orbit in given as:

v = sqrt[(Ge*M)/R],

where Ge is the gravitational constant (Ge = 6.673 x 10^-11N/m2/kg2), M is the mass of the earth(m = 5.98 x 10^24kg), and R is the radius of the earth (R = 6.47 x 10^6m)

v = SQRT [ (6.673 x 10^-11 N m2/kg2) • (5.98 x 10^24 kg) / (6.47 x 10^6 m) ]

v = 7.85 x 10^3 m/s

Explanation:

For a satelite in a low altitude orbit around the Earth, the gravitational force is the only force acting of the said satellite keeping it is a circular orbit. To keep this satellite in perfect circular orbit, it must be moving in at a certain speed, which is dependent on the earth mass and radius. This speed can be evaluated from the expression of centripetal force(F = mv2/r). The centripetal force Fc on the satellite is equal to the gravitational force on the satellite from the earth(Fe). That is, (Ge*M*m)/R2 = (m*v2)/R, where M is mass of the earth, and m is the mass of the satellite. making v the subject of the formula, the equation become v = sqrt[(Ge*M)/R].

hi there if you are asking for conformation that wavelingth is the correct answer is wrong it is the answer is frequency. : )

i hope this you out, have a wonderful day

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6