Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
Tangential speed vT, for the stallite that orbits the earth from a distance 3R is compared by [tex]\bold {(vT)= \dfrac {(v0)}{3}}[/tex].
Given here,
Vo - Speed
R - Distance of center of the earth
Fo - Force exerted on satellite
vT - Tangential speed
For satelite, Gravitational force (Fg)is equal to the centripetal force(Fc),
Fg = Fc = Fo
[tex]\bold {\dfrac {GMm}{R^2} = \dfrac {m(v0)^2}R}[/tex]
GM = (v0)²R..................... (1)
Since, For satellite R = 3R
Thus,
[tex]\bold {\dfrac {GMm}{3R^2} = \dfrac {m(v0)^2}{3R}}[/tex]
[tex]\bold {\dfrac {GMm}{3R^2} = \dfrac {m(vT)^2}{3R}}\\\\\bold {\dfrac {GM}{27 R} = \dfrac {(vT)^2}{3R}}\\\\\bold {GM = \dfrac {27 R \times (vT)^2}{3R}}\\\\\bold {{GM}= 9 R \times (vT)^2}\\[/tex].........(2)
From equation 1 and 2,
(v0)²R = 9(vT)²R
Divide through R and then 9, we get
[tex]\bold {(vT)^2 = \dfrac {(v0)^2}{9}}[/tex]
Take the square root,
[tex]\bold {(vT)= \dfrac {(v0)}{3}}[/tex]
Therefore, tangential speed, vT, for the stallite that orbits the earth from a distance 3R is compared by [tex]\bold {(vT)= \dfrac {(v0)}{3}}[/tex].
To know more about tangential speed,
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