Answer :
Answer:
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].
The speed required to remain in the circular orbit is [tex]v = 7.85 \times10^3 m/s[/tex]
Gravitational Force and Circular motion:
The gravitational force of the earth provides the necessary acceleration to keep the satellite in orbit. It acts as a centripetal force and the acceleration is always towards the center, keeping the satellite in a circular motion.
Let the mass of earth be M and the mass of the satellite be m, and the radiius of the earth be R
The equation of motion, in this case, can be written as:
[tex]\frac{GMm}{R}=\frac{mv^2}{R}[/tex]
[tex]v = \sqrt{\frac{GM}{R}}[/tex]
where G is the gravitational constant ([tex]G = 6.673 \times 10^{-11}\frac{N}{kg^2m^2}[/tex]),
M is the mass of the earth([tex]m = 5.98 \times 10^24kg[/tex]),
and R is the radius of the earth ([tex]R = 6.47 \times 10^6m[/tex])
[tex]v = \sqrt{\frac{ 6.673 \times 10^{-11} \times5.98 \times 10^{24} } {6.47 \times 10^6 } }[/tex]
[tex]v = 7.85 \times10^3 m/s[/tex]
Learn more about circular motion:
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