Calculate the velocity with which a body must be thrown vertically from the surface of the earth so that it may reach a height of `10R`, where `R` is the radus of the earth and is equal to `6.4xx10^(8)m`. (earth's mass`=6xx10^(24)kg`, gravitational constant `G=6.7xx10^(-11)Nm^(2)kg^(-2))`
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The gravitational potential 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 its 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 height `10R` from earth's surface, i.e, at a distance `11R` from earth's centre is `U(11R)=-(GMm)/R` Therefore, 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 given to the body in sending it to that height. Now, suppose the body is throuwn up with a vertical speed `v`, so that its initial kinetic energy is `1/2mv^(2)`. Then `1/2mv^(2)=10/11(GMm)/R` or `v=sqrt(20/11 (Gmm)/R)` putting the given values: `v=sqrt((20xx(6.7x10^(-11)Nm^(2)kg^(-2))xx(6xx10^(24)kg))/(11(6.4xx10^(6)m))` `=1.07xx10^(4)ms^(-1)`
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