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A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on the satellite is F0. An identical satellite orbits Earth at a distance of 3R from the center of the Earth. How does the tangential speed, vT, of the satellite at distance 3R compare to the speed v0 of the satellite at R

Sagot :

oladayoademosu

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,

https://brainly.com/question/17689540

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