A planet has mass ${\frac{1}{10}}^{t h}$ of that of earth, while radius is ${\frac{1}{3}}^{r d}$ that of earth. If a person can throw a stone on earth surface to a height of 90 m, then he will be able to throw the stone on that planet to a height (in meter)-

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Solution

$g = \frac{G M}{R^{2}}$
$\frac{g_{p}}{g_{e}} = \frac{M_{P}}{M_{e}} {(\frac{R_{e}}{R_{p}})}^{2}$
$= \frac{1}{10} \times {(\frac{3}{1})}^{2} = \frac{9}{10}$
$H = \frac{u^{2}}{2 g}$
$\frac{H_{p}}{H_{e}} = \frac{g_{e}}{g_{p}}$
$H_{p} = \frac{10}{9} \times 90 = 100 m$

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A planet has mass 110th of that of earth, while radius is 13rd that of earth. If a person can throw a stone on earth surface to a height of 90 m, then he will be able to throw the stone on that planet to a height (in meter)-

g=GMR2 gpge=MPMe(ReRp)2 =110×(31)2=910 H=u22g HpHe=gegp Hp=109×90=100 m

A planet has mass ${\frac{1}{10}}^{t h}$ of that of earth, while radius is ${\frac{1}{3}}^{r d}$ that of earth. If a person can throw a stone on earth surface to a height of 90 m, then he will be able to throw the stone on that planet to a height (in meter)-

$g = \frac{G M}{R^{2}}$
$\frac{g_{p}}{g_{e}} = \frac{M_{P}}{M_{e}} {(\frac{R_{e}}{R_{p}})}^{2}$
$= \frac{1}{10} \times {(\frac{3}{1})}^{2} = \frac{9}{10}$
$H = \frac{u^{2}}{2 g}$
$\frac{H_{p}}{H_{e}} = \frac{g_{e}}{g_{p}}$
$H_{p} = \frac{10}{9} \times 90 = 100 m$