The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets
Solution 1
Escape velocity of a projectile from the Earth, vesc = 11.2 km/s
Projection velocity of the projectile, vp = 3vesc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = vf
Total energy of the projectile on the Earth = `1/2mv_p^2 - 1/2mv_"esc"^2`
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = `1/2mv_f^2`
From the law of conservation of energy, we have
`1/2mv_p^2 - 1/2mv_"esc"^2 = 1/2 mv_f^2`
`v_f = sqrt(v_p^2-v_"esc"^2)`
`=sqrt((3v_"esc")^2 - (v_"esc")^2`
`=sqrt8 v_"esc"`
`=sqrt8 xx11.2 = 31.68` km/s
Solution 2
Let v_"es" be the escape speed from surfce of Earth havinf a vlaue `v_"es"` = 11.2 kg `s^(-1)`
=`11.2 xx 10^3 ms^(-1)`
By definition
`1/2 mv_e^2 = (GMm)/(R^2)`
When a body is projected with aspeed `v_i = 3v_"es" = 3 xx 11.2 xx 10^3` m/s then it will have a final spee `v_f` such that
`1/2 mv_f^2 = 1/2mv_i^2 - (GMm)/R^2 = 1/2mv_i^2 - 1/2mv_e^2`
=>`v_f = sqrt(v_i^2 -v_e^2)`
`=sqrt((3xx11.2xx10^3)-(11.2xx10^3)^2)`
= `11.2 xx 10^3 xx sqrt8`
=`31.7xx10^3 ms^(-1)` or 31.7 km `s^(-1)`
