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प्रश्न
A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2 R where R= 6400 km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R ?
[G = 6.67 × 10–11 SI units and M = 6 × 1024 kg]
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उत्तर
According to the diagram,
rp = radius of perigee = 2R
ra = radius of perigee = 6R
a = semi-major axis of the ellipse
Hence, we can write
`r_a = a(1 + e) = 6R`
`r_p = a(1 - e) = 2R`
`(a(1 + e))/(a(1 - e)) = (6R)/(2R)` = 3
By solving, we get eccentricity `e = 1/2`
If va and vp are the velocities of the satellite (of mass m) at aphelion and perihelion respectively, then by conservation of angular momentum
`L_("at perigee") = L_("at apogee")`
∴ `mv_pr_p = mv_ar_a`
∴ `v_a/v_p = r_p/r_a = 1/3`
Applying conservation of energy,
The energy at perigee = Energy at apogee
`1/2 mv_p^2 - (GMm)/r_p = 1/2 mv_a^2 - (GMm)/r_a`
Where M is the mass of the earth
∴ `v_p^2 (1 - 1/9) = - 2GM (1/r_a - 1/r_p)`
= `2GM(1/r_p - 1/r_a)` .....(By putting `v_a = v_p/3`)
`v_p = [2GM(1/r_p - 1/r_a)]^(1/2)/([1 - (v_a/v_p)^2]^(1/2)`
= `[((2GM)/R (1/2 - 1/6))/((1 - 1/9))]^(1/2)`
= `((2/3)/(8/9) (GM)/R)^(1/2)`
= `sqrt(3/4 (GM)/R)`
= 6.85 km/s
vp = 6.885 km/s, va = 2.28 km/s
For circular orbit of radius r,
vc = orbital velocity = `sqrt((GM)/r)`
For r = 6R, vc = `sqrt((GM)/(6R)` = 3.23 km/s.
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