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Answer 1 · 2017-10-06T05:13:20

$2114" km"$

$F = G.m.M/(r^2)$

$color(red)(g=10m/s^2$

$color(green)(M = 6times10^24kg$ (mass of the Earth)

$color(blue)(G = 6.7times10^-11 Nm^2kg^-2$

$color(orange)(m = 50kg$

$color(violet)(r=6.4*10^6m = 6341km$ (Radius of earth)

Weight of $m=mg=50kg times10m/s^2=color(indigo)(500N$

We need to find $r'$ such that $"Weight"= color(purple)(250N$
(half the weight of a 50kg mass)

$F = (G*m*M)/(r^2)$

$=>color(purple)(250N) = (color(blue)(6*10^-11)*color(green)(6*10^24)*color(orange)(50))/(r')^2= (1.8*10^16)/(r')^2$

$=>(r')^2=(1.8*10^16)/(250)=color(grey)(7.2*10^30$

$=>r'=sqrtcolor(grey)(7.2*10^30)~~8485281m~~color(blue)(8485km$

So, distance from earth's surface $=r'-r$
$=color(blue)(8485km-color(violet)(6341km))~~2114km$