Calculate the velocity with which a body must be thrown vertically upward from the surface of the earth so that it may reach a height of `10R`, where `R` is the radius of the earth and is equal to `6.4 xx 10^(6)m.` (Given: Mass of the earth `= 6 xx 10^(24) kg`, gravitational constant `G = 6.7 xx 10^(-11) N m^(2) kg^(-2)`)

Explanation

The gravitational potential energy of a body of mass m on earth s surface is
`U (R) = - (GMm)/(R)`
where `M` is the mass of the earth (supposed to be concentrated at itss centre) and `R` is the radius of the earth (distance of the particle from the centre of the earth) The gravitational energy of the same body at a heightr `10R` from eqrth's surface i.e at a distance `11R` from earth's centre is
`U (11R) = - (GMm)/(R)`
`:.` change in potential energy
`U (11R) - U (R) = - (GMm)/(11R) - (-(GMm)/(R)) = (10)/(11) (GMm)/(R)`
This difference must come from the initial kinetic energy givento the body in sending it to that height Now suppose the body is thrown up with a vertical speed so that its initial kinetic energy is `(1)/(2) mv^(2)` Then
`(1)/(2) mv^(2) = (10)/(11) (GMm)/(R)` or `v = sqrt((20/11(GMm)/R)`
Putting the given values `v=sqrt(((20xx(6.7xx10^(11)N-m^(2)//kg^(2))xx(6xx10^(24)kg))/(11(6.4xx10^(6)m))))=1.07xx10^(4)m//s` .