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A body is projected vertically up with a speed $\mathrm{v_0=\sqrt{g R}}$ from earth's surface where $\mathrm{\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^2}$ and $\mathrm{\mathrm{R}=}$ radius of earth $=6.4 \times 10^6 \mathrm{~m}$ . What is the maximum height attained by the body.

Option: 1
$6400 \mathrm{Km}$

Option: 2
$36000 \mathrm{Km}$

Option: 3
$42400 \mathrm{Km}$

Option: 4
$29600 \mathrm{Km}$

Answers (1)

best_answer

Let the body of mass $\mathrm{m}$ be projected from point 1 and attains a height $\mathrm{h}$ at point 2. Conservation of energy of the body between 1 and 2 yields, $\mathrm{\mathrm{U}_1+\mathrm{K}_1=\mathrm{U}_2+\mathrm{K}_2}$

$\mathrm{ \Rightarrow\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)+\left(\frac{1}{2} \mathrm{mv}_1^2\right)=\left(-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\right)+\left(\frac{1}{2} \mathrm{mv}_2^2\right) }$

Putting $\mathrm{ \mathrm{v}_2=0\: and \: \mathrm{v}_1=\mathrm{v}_0 }$ , we obtain,

$\mathrm{\frac{1}{2} m_0^2=G M m\left(\frac{1}{R}-\frac{1}{R+h}\right)}$

$\mathrm{\Rightarrow \frac{v_0^2}{2}=\frac{G M h}{R(R+h)} }$

$\mathrm{\Rightarrow \frac{v_0^2}{2}=\left(\frac{G M}{R^2}\right)\left(\frac{R h}{R+h}\right) }$

$\mathrm{\Rightarrow \frac{v_0^2}{2}=\frac{g R h}{R+h} \quad\left(\because \frac{G M}{R^2}=g\right) }$

$\mathrm{ \frac{R+h}{h}=\frac{2 g R}{v_0^2} }$

$\mathrm{ \Rightarrow \frac{R}{h}+1=\frac{2 g R}{v_0^2} }$

$\mathrm{\Rightarrow h=\frac{R}{\frac{2 g R}{v_0^2}-1}}$

Putting $\mathrm{v_0=\sqrt{g R},}$ we obtain,

$\mathrm{\Rightarrow \mathrm{h}=\mathrm{R}=6400 \mathrm{~km}}$

Hence option 1 is correct.

Posted by

Nehul

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