Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to (3/4) th of the escape velocity from the surface of the earth? (R a d i u s o f t h e e a r t h = R) :

Question

Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to (3/4) th of the escape velocity from the surface of the earth? (R a d i u s o f t h e e a r t h = R) : (A) (10 R/9) (B) (16 R/7) (C) (9 R/8) (D) (10 R/3)

Tardigrade Admin · Accepted Answer

v=(3/4) ve K.E.=(1/2) (m v)2=(1/2)m ((3/4) ve)2 =(9/32) mve2 =(9/32)m ((2 G M/R)) K.E=(9/16) (G M m/R) P.E.=-(G M m/R) Total energy=K.E.+P.E.=-(7/16) (G M m/R) Let the distance from the centre of earth be h ; then P.E.=-(G M m/h) Total energy=P.E. above earth's surface -(7/16)(G M m/R)=- (G M m/h) ∴ h=(16 R/7)

Q. Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to $\frac{3}{4}$ th of the escape velocity from the surface of the earth? $(R a d i u s o f t h e e a r t h = R)$ :

$\frac{10 R}{9}$

$\frac{16 R}{7}$

$\frac{9 R}{8}$

$\frac{10 R}{3}$

Q. Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to 43​ th of the escape velocity from the surface of the earth? (Radiusoftheearth=R) :

910R​

716R​

89R​

310R​

Q. Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to $\frac{3}{4}$ th of the escape velocity from the surface of the earth? $(R a d i u s o f t h e e a r t h = R)$ :

$\frac{10 R}{9}$

$\frac{16 R}{7}$

$\frac{9 R}{8}$

$\frac{10 R}{3}$