Question

Question #159c0

Answer 1

`W=1/3(GMm)/R=(mgR)/3`

Explanation:

Gravitational potential energy of mass `m`
`U=-(GMm)/r`
`G=` Universal gravitational constant
`M=`mass of earth
`m=` mass of object
`r=`distance between center of Earth and center of object

Gravitational potential energy on the surface of earth `(r=R)`
`U_1=-(GMm)/R`

Gravitational potential energy on the height `R/2` rom the surface of earth `(r=R+R/2=3R/2)`
`U_2=-(GMm)/((3R)/2)=-2/3(GMm)/R`
work done to send a body of mass m from earth surface to height R/2
`="change in potential energy of object"`
`W=DeltaU=U_2-U_1`
`W=-2/3(GMm)/R-(-(GMm)/R)=-2/3(GMm)/R+(GMm)/R`
`W=(GMm)/R (1-2/3)`
`W=1/3(GMm)/R`

We know that relation between Universal gravitational constant and acceleration due to gravity`g` is `GM=gR^2`

`W=1/3(gR^2m)/R=(mgR)/3`