Question 13 of 32 1 Point For a satellite to orbit Earth at a constant distance, its centrifugal acceleration must be equa

Question

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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3 years 2021-08-27T12:00:38+00:00 1 Answers 1 views 0

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