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???

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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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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

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

Approved