__Question 8.17__:^{–1}from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10

^{24}kg; mean radius of the earth = 6.4 × 10

^{6}m; G= 6.67 × 10

^{–11}N m

^{2}kg

^{–}

^{2.}

__Solution__:*v*= 5 km/s = 5 × 10

^{3}m/s

Mass of the Earth, M

_{e}= 6 X 10

^{24}kg

Radius of the Earth, R

_{e}= 6.4 X 10

^{6}m

Height reached by rocket mass,

*m*=

*h*

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

= (1/2)mv

^{2}+ (-GM

_{e}m / R

_{e})

At highest point

*h*,

v = 0

And, Potential energy = -GM

_{e}m / (R

_{e}+ h)

Total energy of the rocket = 0 + [ -GM

_{e}m / (R

_{e}+ h) ]

= -GM

_{e}m / (R

_{e}+ h)

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height

*h*

(1/2)mv

^{2}+ (-GM

_{e}m / R

_{e}) = -GM

_{e}m / (R

_{e}+ h)

(1/2)v

^{2}= GM

_{e}[ (1/R

_{e}) - 1 / (R

_{e}+ h) ]

= GM

_{e}[ (R

_{e}+ h - R

_{e}) / R

_{e}(R

_{e}+ h) ]

(1/2)v

^{2}= gR

_{e}h / (R

_{e}+ h)

Where g = GM / R

_{e}

^{2}= 9.8 ms

^{-2}

∴ v

^{2}(R

_{e}+ h) = 2gR

_{e}h

_{}

v

^{2}R

_{e}= h(2gR

_{e}- v

^{2})

h = R

_{e}v

^{2}/ (2gR

_{e}- v

^{2})

_{}

= 6.4 X 10

^{6}X (5 X 10

^{3})

^{2}/ [ 2 X 9.8 X 6.4 X 10

^{6}- (5 X 10

^{3})

^{2}

h = 1.6 X 10

^{6}m

Height achieved by the rocket with respect to the centre of the Earth = R

_{e}+ h

= 6.4 X 10

^{6}+ 1.6 X 10

^{6}= 8 X 10

^{6}m

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