Question

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A rocket is fired 'vertically' from the surface of mars with a speed of 2 km s^(-1). If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×10²³ kg; radius of mars = 3395 km; G = 6.67×10^-11 N m² kg^-2 .
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Answer 1

Answer:

Given that:-

• Initial velocity of the rocket (v) = 2km/s

In m/s :-

2km/s = 2×10³ m/s

• Mass of Mars (M) = 6.4 × 10²³ kg
• Radius of Mars (R) = 3395 km

In m :-

3395 km = 3.395×10⁶ m

• Universal gravitational constant ;

$\mapsto \: G = 6.67 \times {10}^{ - 11} \: N \: {m}^{2} \: {kg}^{ - 2}$

Let :-

• Mass of Rocket = m

★ Initial kinetic energy of the rocket :-

=> 1/2 mv²

★ Initial potential energy of the rocket :-

$\mapsto \: \frac{ - GMm}{R}$

Therefore,

★ Total Initial energy :

=> Initial kinetic energy - Initial potential energy

$\longmapsto \: \frac{1}{2} {mv }^{2} \: - \frac{GMm}{R}$

To Find :-

• If 20% of its initial energy is lost due to Martian atmospheric resistance, then only 80% of its kinetic energy in reaching a height. So, How far will the rocket go from the surface of mars before returning to it?

Solution :-

Total initial energy available :-

$\mapsto \: \frac{80}{100} \times \frac{1}{2} {mv}^{2} - \frac{GMm}{R} = 0.4 {mv }^{2} - \frac{GMm}{R}$

Let, The maximum height reached by the rocket = 'h'.

___________________

At this height, the velocity and hence, the kinetic energy of the rocket will become zero(0).

★ Total energy of rocket height (h) :-

$\mapsto \: - \frac{ GMm}{(R + h)}$

_____________________

Applying the laws of conversation of energy for the rocket, we can write:

$0.4 {mv}^{2} - \frac{GMm}{R} = \frac{ - GMm}{(R + h)}$

$0.4 {v}^{2} = \frac{GM}{R} - \frac{GM}{R + h}$

$\mapsto \: Gm( \frac{1}{R} - \frac{1}{R + h} )$

$\mapsto \: Gm \: ( \frac{R + h -R}{R(R + h)} )$

$\mapsto \: \frac{GMh}{R(R + h)}$

$\mapsto \: \frac{R + h}{h} = \frac{GM}{ {0.4v}^{2}R }$

$\mapsto \: \frac{R}{h} + 1 = \frac{GM}{ {0.4v}^{2}R }$

$\mapsto \: \frac{R}{h} = \frac{GM}{ {0.4v}^{2}R } - 1$

$\mapsto \: \frac{R}{h} = \frac{R}{ \frac{GM}{ {0.4v}^{2}R } - 1 }$

$\mapsto \: \frac{ {0.4R}^{2} {v}^{2} }{GM - {0.4v}^{2}R }$

$\mapsto \: \frac{0.4 \times (3.395 \times {10}^{6})^{2} \times ( {2 \times 10}^{3})^{2}}{6.64 \times {10}^{ - 11} \times 6.4 \times {10}^{23} - 0.4 \times { ({2 \times 10}^{3} )}^{2} \times (3.395 \times {10}^{6}) }$

$\mapsto \: \frac{18.442 \times {10}^{11} }{42.688 \times {10}^{12} - 53432 \times {10}^{12} } = \frac{18.442}{37.256} \times {10}^{6}$

$\mapsto \: 495 \times {10}^{3} m= 495km$

Answer 2

Answer:

$\large\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

Step-by-step explanation:

Let m be the mass of the rocket,

$\[ \begin{aligned} \tt M & \rm=\text { mass of Mars }=6.4 \times 10^{23} kg , \\ \\ \tt R & \rm=\text { radius of Mars }=3395 km , \\ \\ \tt v & \rm=\text { initial speed of rocket }=2 \times 10^{3} m s ^{-1}, \end{aligned} \]$

h= maximum height reached by the rocket.

Initial kinetic energy of the rocket $\rm=\dfrac{1}{2} m v^{2} ,$

initial potential energy of the rocket $\rm=-\dfrac{G M m}{R} ,$

and energy lost due to Martian atmosphere $\rm=\dfrac{20}{100} \times \dfrac{1}{2}{m v}^{2}=\dfrac{1}{10}{m v}^{2}$

From energy conservation principle, total initial mechanical energy = energy loss due to resistance + remaining potential energy.

$\[ \begin{array}{ll} \displaystyle\rm \therefore \frac{1}{2} m v^{2}+\left(\frac{-G M m}{R}\right)=\frac{1}{10} m v^{2}+\left(-\frac{G M m}{R+h}\right) \\\\ \displaystyle\rm \text { or } G M m\left(\frac{1}{R}-\frac{1}{R+h}\right)=m v^{2}\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{2}{5} m v^{2} \\ \\ \displaystyle\rm\text { or } \quad \frac{h}{R(R+h)}=\frac{2 v^{2}}{5 G M} \\ \\ \displaystyle\rm\text { or } \quad \frac{R+h}{h}=\frac{5 G M}{2 R v^{2}} \\ \\ \displaystyle\rm\text { or } \quad \frac{R}{h}=\frac{5 G M}{2 R v^{2}}-1=\frac{5 G M-2 R v^{2}}{2 R v^{2}} \\ \\ \displaystyle\rm\Rightarrow \quad \text { required height } h=\frac{2 R^{2} v^{2}}{5 G M-2 R v^{2}}=\frac{0.4 R^{2} v^{2}}{G M-0.4 R v^{2}} \\\\ \displaystyle\rm \quad=\frac{0.4\left(3395 \times 10^{3} m \right)^{2}\left(2000 m s ^{-1}\right)^{2}}{\left(6.67 \times 10^{-11} N m ^{2} kg ^{-2}\right)\left(6.4 \times 10^{23} kg \right)-0.4\left(3395 \times 10^{3} m \right)\left(2000 m s ^{-1}\right)^{2}} \\\\ \displaystyle\rm \quad=\frac{18.44 \times 10^{18} m ^{4} s ^{-2}}{\left(42.688 \times 10^{12}-5.432 \times 10^{12}\right) m ^{3} s ^{-2}}\\ \\ \\ \displaystyle\rm=0.495 \times 10^{6} m \\ \\ \\ \boxed{\color{violet}\displaystyle\rm =495 km .} \end{array} \]$