 # 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???

11 years ago

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.)

All the best.

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Sagar Singh

B.Tech, IIT Delhi

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11 years ago

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