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Byju's Answer

Standard XII

Physics

Escape Speed

A particle is...

Question

A particle is projected vertically upward from the surface of the earth with a speed of $\sqrt{\frac{3}{2} g R}$ , $R$ being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

\frac{3 R}{4}

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R

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2 R

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3 R

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Solution

The correct option is D $3 R$
$\frac{1}{2} m v_{0}^{2} - \frac{G M m}{R}$
$= - \frac{G M m}{R + h}$ and $g = \frac{G M}{R^{2}}$
Therefore $\frac{1}{2} \times \frac{3}{2} \times gR - Rg = - \frac{R^{2} g}{R + h}$
$- \frac{R}{4} = - \frac{R^{2}}{R + h}$
$R + h = 4 R$
We get $h = 3 R$

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A particle is projected vertically upward from the surface of the earth with a speed of √32gR, R being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

The correct option is D 3R12mv20−GMmR =−GMmR+h and g=GMR2 Therefore 12×32×gR - Rg=−R2gR+h −R4=−R2R+h R+h=4R We get h=3R

A particle is projected vertically upward from the surface of the earth with a speed of $\sqrt{\frac{3}{2} g R}$ , $R$ being the radius of the earth and g is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)

The correct option is D $3 R$
$\frac{1}{2} m v_{0}^{2} - \frac{G M m}{R}$
$= - \frac{G M m}{R + h}$ and $g = \frac{G M}{R^{2}}$
Therefore $\frac{1}{2} \times \frac{3}{2} \times gR - Rg = - \frac{R^{2} g}{R + h}$
$- \frac{R}{4} = - \frac{R^{2}}{R + h}$
$R + h = 4 R$
We get $h = 3 R$