Question

A satellite is revolving around the earth at height h the satellite explodes into two equal parts one part has zero velocity after the explosion and the other moves in the same direction with double the orbital speed of the original satellite the maximum height reached by the part with double the velocity is

Answer 1

let initial the satellite has mass M and mass of earth is m ,

initially the satellite is at distance r= h+R (from the centre of earth)and when it reaches the maximum height its distance is r0 from centre of earth hence r0 =R+H (here H is height from surface of earth)

applying angular momentum conservation principle after breaking and when itb reaches its max height

m1 v0r1=m2 v1r2 (here v0and v1 are orbital speed)

2M×v0×r =M×v1×r0 v1=2v0rr0....1>

now initial total energy after breaking and when it reaches the maximum height are equal

Ui +Ki

=Uf +Kf−2GMmr+12M(2v0)2

= −GMmr0+12M(v1)2

now putting the value of v1 from eq 1 we get

−2GMmr+12M(2v0)2

=−GMmr0+12M(2v0rr0)2−2Gmr+2v02

=−Gmr0+2v20r2r02 ....2>now formula of orbital speed

v0=Gmr−−−√

and putting in eq 2 and solving above equation

r0 =2r=2R+2hR+H

=2R+2hH

=R+2h