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A satellite is in an elliptical orbit around the Earth with aphelion of 6R and perihelion of 2R 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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Answered by ajaydhayal

Answer:

A satellite is in an elliptical orbit around the Earth with aphelion of 6R and perihelion of 2R 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 ?

rp=2R,ra=6R

rp=2R,ra=6RHence rp=a(1−e)=2R(i)

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii)

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii) 1+e1−e=62

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii) 1+e1−e=623−3e=1+e

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii) 1+e1−e=623−3e=1+e4e=2⇒e=21

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii) 1+e1−e=623−3e=1+e4e=2⇒e=21There is not external force or torque on system.

rp=2R,ra=6RHence rp=a(1−e)=2R(i)ra=a(1+e)=6R(ii)On dividing (i) by (ii) 1+e1−e=623−3e=1+e4e=2⇒e=21There is not external force or torque on system.So by the law of conservation of angular momentum.

mavara=mpvprpma=mp=m= mass of satellite

mavara=mpvprpma=mp=m= mass of satellite ∴vpva=rarp6R2R=31

mavara=mpvprpma=mp=m= mass of satellite ∴vpva=rarp6R2R=31 So, Vp=3Va

mavara=mpvprpma=mp=m= mass of satellite ∴vpva=rarp6R2R=31 So, Vp=3VaApply conservation of energy at apogee an d perigee

mavara=mpvprpma=mp=m= mass of satellite ∴vpva=rarp6R2R=31 So, Vp=3VaApply conservation of energy at apogee an d perigee21mvp2−rpGMm=21mva2−raGMm

mavara=mpvprpma=mp=m= mass of satellite ∴vpva=rarp6R2R=31 So, Vp=3VaApply conservation of energy at apogee an d perigee21mvp2−rpGMm=21mva2−raGMmMultiplying m2 to both side and putting rp=2R and r

Va=3Vp

Va=3Vp∴vp2−va2=RGM−31RGM

Va=3Vp∴vp2−va2=RGM−31RGMvp2−(3vp)2=RGM[1−31]

Va=3Vp∴vp2−va2=RGM−31RGMvp2−(3vp)2=RGM[1−31]vp2[1−91]=RGM32

Va=3Vp∴vp2−va2=RGM−31RGMvp2−(3vp)2=RGM[1−31]vp2[1−91]=RGM32vp298=RGM.32

Va=3Vp∴vp2−va2=RGM−31RGMvp2−(3vp)2=RGM[1−31]vp2[1−91]=RGM32vp298=RGM.32vp2=R

=6.85×103m/s=6.85km/s

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/s

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/svc=rGM=6R6.67×10−11×6×1024

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/svc=rGM=6R6.67×10−11×6×1024=6×6.4×1066.67×6×1024−11=640667×10613−5

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/svc=rGM=6R6.67×10−11×6×1024=6×6.4×1066.67×6×1024−11=640667×10613−5=1.042×10×105=10.42×106

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/svc=rGM=6R6.67×10−11×6×1024=6×6.4×1066.67×6×1024−11=640667×10613−5=1.042×10×105=10.42×106Vc=3.23km/s

=6.85×103m/s=6.85km/sva=3vp=36.85=2.28km/svc=rGM=6R6.67×10−11×6×1024=6×6.4×1066.67×6×1024−11=640667×10613−5=1.042×10×105=10.42×106Vc=3.23km/shence to transfer to a circular orbit at apogee we have to boost the velocity by

v0−va=(3.23−2.28)=0.95km/s

Explanation:

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