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Question 13 of 32 1 Point For a satellite to orbit Earth at a constant distance, its centrifugal acceleration must be equa
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Question 13 of 32
1 Point
For a satellite to orbit Earth at a constant distance, its centrifugal acceleration
must be equal and opposite Earth’s gravitational acceleration. If a satellite is
to orbit at a constant distance from Earth at a circular radius of 7,000,000 m,
what is the required velocity of the satellite? (Assume the acceleration due to
Earth’s gravity is 8.2 m/s2 at this altitude.)
O A. 8239 m/s
O B. 7576 m/s
C. 6818 m/s
D. 7043 m/s
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Physics
3 years
2021-08-27T12:00:38+00:00
2021-08-27T12:00:38+00:00 1 Answers
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Answer: B
V = 7576 m/s
Explanation: Given that the
circular radius R = 7,000,000 m,
The acceleration due to Earth’s gravity is 8.2 m/s2
Let’s use gravitational field strength formula to get the mass of the satellite.
g = Gm/R^2
Where G = 6.674×10^−11m3⋅kg−1⋅s−2
Substitutes all the parameters into the formula
8.2 = (6.674×10^-11 × M)/(7000000)^2
Cross multiply
4.018 × 10^14 = 6.674×10^−11M
Make M the subject of formula
M = 4.018 × 10^14 ÷ 6.674×10^−11
M = 6.02 × 10^24 kg
The Force of attraction experienced by the satellite will be
F = mg
F = 6.02 × 10^24 × 8.2
F = 4.94 × 10^25 N
But centripetal force is
F = MV^2/ R
Where V = velocity of the satellite
Substitute F, M and R into the formula
4.94×10^25 = (6.02×10^24V^2)/ 7 ×10^6
Cross multiply
3.46 × 10^32 = 6.02×10^24V^2
Make V^2 the subject of formula
V^2 = 3.46×10^32 ÷ 6.02×10^24
V^2 = 57403600.2
Find the square root of the value to get V
V = 7576 m/s
The required velocity of the satellite is 7576 m/s approximately