The change in the P-E- when a body of mass m is displaced from Earth-s surface to a vertical height equal to radius of earth -g - acceleration due to gravity on earth surface- is-dfrac-mgR-2-dfrac-2mgR-3-dfrac-mgR-3-dfrac-3mgR-4-

Question

Toppr Admin · Accepted Answer

-x394-PE-GMmR-1-x2212-12-GMm2RgR-GMR-x394-PE-mgR2

The change in the P.E., when a body of mass $m$ is displaced from Earth's surface to a vertical height equal to radius of earth (g $=$ acceleration due to gravity on earth surface) is:
$\frac{m g R}{2}$
$\frac{2 m g R}{3}$
$\frac{m g R}{3}$
$\frac{3 m g R}{4}$

$Δ P E = \frac{G M m}{R} {1 - \frac{1}{2}} = \frac{G M m}{2 R}$
$g R = \frac{G M}{R}$
$Δ P E = \frac{m g R}{2}$

The change in the P.E., when a body of mass m is displaced from Earth's surface to a vertical height equal to radius of earth (g = acceleration due to gravity on earth surface) is:mgR22mgR3mgR33mgR4

ΔPE=GMmR{1−12}=GMm2RgR=GMRΔPE=mgR2

The change in the P.E., when a body of mass $m$ is displaced from Earth's surface to a vertical height equal to radius of earth (g $=$ acceleration due to gravity on earth surface) is:
$\frac{m g R}{2}$
$\frac{2 m g R}{3}$
$\frac{m g R}{3}$
$\frac{3 m g R}{4}$

$Δ P E = \frac{G M m}{R} {1 - \frac{1}{2}} = \frac{G M m}{2 R}$
$g R = \frac{G M}{R}$
$Δ P E = \frac{m g R}{2}$