A particle is fired vertically upward fom earth's surface and it goes up to a maximum height of 6400 km. find the initial speed of particle.

Answer

A

Explanation

The particle attain maximum height =6400 km
On earth's surface its P.E. and K.E.
`E_e=1/2MV^2+((-gmM)/r)`…….. 1
In space its P.E. and K.E.
`E_3=((-GMm)/(R+h))+0`
`E_3=((-GMm)/(2R))`……….2 `(:.g=R)`
Equation 1 and 2
`=-(GMm)/R+1/2mv^2=-(GMm)/(2R)`
or `(1/2)mv^2=GMm(-1/(2R)+1/R)`
`v^2=(GM)/R`
`=(6.67xx10^-11xx6xx10^24)/(6400xx10^3)`
`(40.02+10^13)/(6.4xx10^6)`
`=6.2xx10^7=0.62x10^8`
or `v=sqrt(0.62xx10^8)`
`=0.79xx10^4m/s`
`=79km/s`