Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
Learn more about NEO here:
https://brainly.com/question/24157038
