11PE = G*m*dr/R2
PE = -Gm/R
Work To Be done = Gm/R
The minimum energy required to launch a m kg satellite from earth;s surface in a circular orbit at an altitude 2R where R is the radius of earth
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5 Answers
1no virang , u r wrong
let me give u options
a)5mgR/3
b)4mgR/3
c)5mgR/6
d)5mgR/4
3net energy on surface of earth ... -GMm/R
PE on the other place = -GMm/3R [CONCIDERING ALT FRM SURFACE OF EARTH]
mV^2/3R = GMm/(3R)^2
V^2 = GM/3R
1/2 m v^2 = GMm/6R ..........
GMm/6R -GMm/3R
-GMm/6R......
ENGY SUPPLIED
-GMm/6R+GMm/R = 5GMm/6R = 5mgR/6 .......OPTION C .......
[1]
1Energy to be supplied = Final energy - Initial energy
Initial state --> Only PE is present.
Thus, initial energy =\frac{-GmM }{R}}
Final state --> Both KE and PE are present
Thus, final energy = \frac{-GmM}{2(3R)} = \frac{-GmM }{6R}
Thus, energy to be supplied = \frac{GmM}{R} - \frac{GmM}{6R}= \frac{5GmM }{6R}
Hence, answer is C.
Hope this helps
Please correct me if I'm wrong.
ALL THE BEST