Answer:
[tex]h=25.52\times10^6 m[/tex]
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface
[tex]-G M m / ( R + h )=- G M m / R + (1/2) m v^2[/tex]
[tex]- G M / ( R + h ) = - G M / R + (1/2) v^2[/tex]
[tex]-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2[/tex]
[tex]-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )[/tex]
=[tex]( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2[/tex]
=[tex]-2.50625\times10^7 J[/tex]
[tex]=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7[/tex]
[tex]R+h=31.92\times10^{6}[/tex]
[tex]h=31.92\times10^{6}-6.4\times10^6[/tex]
[tex]h=25.52\times10^6 m[/tex]
