What is the escape velocity of a meteorite situated 1800 km above the surface of the earth? Given, the radius of the earth =6300 km and the acceleration due to gravity on the earth's surface `=9.8m*s^(-2)`.

Explanation

Radius of the earth R =6300 km `=63xx10^5` m
Distance of the meteorite from the centre of the earth r=6300 +1800 =8100 km `=81xx10^5 km`
Acceleration due to gravity on the earth.s surface ,`g=(GM)/(R^2)`
acceleration due to gravity at the position of the meteorite,
`g^.=(GM)/(r^2)`.
`therefore g/(g^.) =(r^2)/(R^2) or, g^.=(gR^2)/(r^2)=g(R/r)^2`
`therefore` Escape velocity of the meteorite,
`v_e^.=sqrt(2g^.r)=sqrt(2g(R^2/r^2)r)=Rsqrt((2g)/r)`
`=63xx10^5xxsqrt((2xx9.8)/(81xx10^5))`
`98xx10^2m*s^(-1)=9.8km*s^(-1)`.