Question

Ques. A rocket is fired vertically with a speed of 5 Km /s from the earth surface . How far from the earth does the rocket go before returning to the earth ? Mass of earth
is 6 × 10²⁴ Kg . Radius = 6.4 × 10^6 .


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Answer 1

→ A rocket is fired vertically with a speed of 5 km/s from the earth's surface . How far from the earth does the rocket go before returning to the earth ?

=> Let us assume that the rocket is fired with an initial velocity 'v' from the earth's surface . Let it reach a maximum height 'h' where its velocity becomes zero.

• Total energy of the rocket will be equal to the sum total of its kinetic and potential energies .

E = K.E + P.E

= 1/2 mv² + [ -GMm/R ]............(Eq.1 )

Now , when it reaches the highest point , the velocity will become zero. Thus, kinetic energy will also become zero. The radius will become ( R + h )

Thus,

• v = 0

• K.E = 0

•P .E = -GMm/(R + h )

◾E = 0 + [ -GMm/(R + h ) ]

=> -GMm/( R + h ).............(Eq.2)

⭐Now , applying the law of conservation of energy :

=> ½mv² - GMm/R = -GMm/(R + h )

{ Take m as a common factor and cancel it out }

=> ½v² = GM/R - GM/ ( R + h )

{ Now substitute GM= gR² }

$= > \frac{g \: r {}^{2} }{r} - \frac{g \: r {}^{2} }{(r + h)}$
{ Take gR as common factor }

$= > gr \: (1 - \frac{r}{(r + h)} )$

Finally we get ,

$= > \frac{1}{2} v {}^{2} = gr \: ( \frac{h}{(r + h)} )$

{ cross multiply the terms and solve }

=> v² ( R + h ) = 2g R h

=> Rv² - v²h = 2g R h

=> Rv² = 2g R h + v²h

( Take h as common in R.H.S )

=> Rv² = h ( 2g R - v² )

→ From this equation we have the value of h =

$= > h \: = \frac{r \: v {}^{2} }{2g \: r \: - v {}^{2} } \\ \\ = > \frac{(6.4 \times 10 {}^{6} ) \times (5 \times 10 {}^{3}) {}^{2} }{2 \times 9.8 \times (6.4 \times 10 {}^{6} ) - (5 \times 10 {}^{3} ){}^{2} }$

( Putting values of mass ( M) , radius ( R ) and Gravitational Constant ( G ) in the equation )

→ Solving this expression we will get

$height \: h \: = 1.6 \times 10 {}^{6} m$

→Now total height achieved = ( Re + h )

=> ( 6.4 × 106 + 1.6 × 10^6 )

=  8 × 10^6 m.

★Hope it's clear !
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Answer 2

Hey sis...

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Here is your answer
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» Given...

Velocity of rocket (v) :- 5 km/s

Mass of earth $(M)$ = 6 × 10²⁴ kg

Radius of earth $(R)$ = 6.4 × ${10}^{6}$

» We have to find the distance covered by the rocket...

Now...

• At the surface of earth...

Total Energy = Kinetic Energy + Potential Energy of the rocket

$\dfrac{1}{2}$mv² + $\dfrac{-GMm}{R}$ ...(1)

• At height...

Kinetic Energy becomes zero

Then

Total Energy = 0 + $[\dfrac{-GMm}{R\:+\:h}]$ ...(2)

According to law of conservation of energy

eq. (1) = eq. (2)

$\dfrac{1}{2}$ mv² + $\dfrac{-GMm}{R}$ = $[\dfrac{-GMm}{R\:+\:h}]$

$\dfrac{1}{2}$ v² = GM $[\dfrac{1}{R}$ - $\dfrac{1}{R\:+\:h}]$

= GM $[\dfrac{R\:+\:h\:-\:R}{R(R\:+\:h}]$

= $\dfrac{GMh}{R(R\:+\:h)}$ × $\dfrac{R}{R}$

Now...

g = $\dfrac{GM}{R}^{2}$

= 9.8 m/s²

So...

$\dfrac{1}{2}$ v² = $\dfrac{gRh}{R\:+\:h}$

v² $R$ = $h(2g R - {v}^{2} )$

h = $\dfrac{R {v}^{2}}{2g R - {v}^{2}}$

Now put all the given values... that we know and that are given in question in above eq.

h = $\dfrac{ {6.4}^{6} \: \times \: (5 \: \times \: {10}^{3})^{2} }{2 \: \times \: 9.8 \: \times \: 6.4 \: \times \: {10}^{6} \: - \: (5 \: \times \: {10}^{3} )^{2} }$

On solving...

h = 1600 km

Now... distance covered by rocket = $R\:+\:h$

= 1600 + 6400

=> 8000 km or 8 × ${10}^{6}$ m

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