Question

An object is thrown with half of the kinetic energy needed to free an object from the attraction of the earth. How far will the object reach from the earth's surface?

Answer 1

`4/3R_E`

Explanation:

Let `v_0` be the initial (launch) velocity of the object that can free itself from the attraction of the earth. It means at maximum high, it's velocity and potential energy there, represented by the the right hand side of the conservation of energy equation below, must be zero.

`1/2mv_0^2 - G(mM_E)/R_E=1/2mv^2 - G(mM_E)/r=0+0=0`
`rArr 1/2cancelmv_0^2 = (cancelmM_E)/R_E^2`
`v_0^2=(2GM_E)/R_E or v_0=sqrt((2GM_E)/R_E`

If another object is launched at half of the `cancel(speed)` kinetic energy above, when it reaches the maximum high at r, it's velocity there is `v=0`. Apply the conservation of energy again:

`½(1/2mv_0^2) - G(mM_E)/R_E=1/2mv^2 - G(mM_E)/r`

`because v=0`,

`1/4cancelmv_0^2 - G(cancelmM_E)/R_E=-G(cancelmM_E)/r`
`1/4v_0^2 - (GM_E)/R_E=-(GM_E)/r`

Substitute `v_0^2=(2GM_E)/R_E` into the equation above,

`1/2(GM_E)/R_E - (GM_E)/R_E=-(GM_E)/r`
`-1/2cancel((GM_E))/R_E=-cancel((GM_E))/r`

`rArr r= 2R_E`

This is from the center of the earth, above the surface of the earth,
`high = 2R_E-R_E=R_E`

Answer 2

Gravitational Potential energy possessed by an object having mass `m` at the surface of earth

`PE=-G(Mm)/R` .......(1)
where `G` is Universal Gravitational constant, `M and R` are mass and radius of earth respectively.

Let `v_e` be the velocity required to free an object from the attraction of the earth.

`KE=1/2mv_e^2` .....(2)

The object will be free when it is placed at `oo`. `:.PE" become "0.` As such work done to move the object to `oo`, must be equal to initial potential energy. Therefore we get

`1/2mv_e^2=G(Mm)/R`
`v_e=sqrt((2GM)/R)` .......(3)

The object is thrown with half the kinetic energy of what we have from (3). Let its velocity `=v`.

`KE=1/2mv^2=1/4mv_e^2`
`=>v=v_e/sqrt2` .....(4)

When the object reaches height `h` above surface of earth its velocity `=0`, thereafter it starts falling towards earth under gravity. Using Law of conservation of energy

`1/2mv^2-G(Mm)/R=1/2m(0)^2-G(Mm)/(R+h)`
`=>1/2v^2-G(M)/R=-G(M)/(R+h)`

Writing first terms using (4) and (3)

`1/4v_e^2-G(M)/R=-G(M)/(R+h)`
`=>1/4(2GM)/R-G(M)/R=-G(M)/(R+h)`
`=>1/2(GM)/R-G(M)/R=-G(M)/(R+h)`
`=>1/(2R)=1/(R+h)`
`=>R+h=2R`
`=>h=R`