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a shell vertically fired from earth with velocity v/n where v is escape velocity and n is some no. greater than 1neglecting the air resistance calculate the max. altitude attained by shell???
Dear student,
Take this hint:
Apply Pythagoras’ theorem to the right-angled triangle to establish that the appropriate speed for a circular orbit just above the earth’s atmosphere is given by
(Use the approximation that the distance traveled in one second is tiny compared to the radius of the earth.)
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Sagar Singh
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in case of conservative forces we can apply conservation of energy ,so
initial total energy (TE)= PE + KE
PE = -GMm/R (at surface)
KE=mv2/2n2
v = (2GM/R)1/2 (escape speed) so
KE = GMm/n2R
so total energy = KE + PE = GMm/n2R - GMm/R ..................1
finally particle reaches to max height (x distance from surface)& its velocity becomes 0 ...
TE = PE + KE
PE = -GMm/(R+x)
KE = 0
TE = -GMm/(R+x) ....................2
since energy is conserved so eq 1 = eq 2
-GMm/(R+x) = -GMm/R + GMm/n2R
X = R/(n2-1) (max altitude)
approve my ans if u like it
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