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7.Gravitation
medium
A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_e\right)$, the increase in potential energy will be : ( $g =$ acceleration due to gravity on the surface of Earth)
A$3 mgR$
B$\frac{1}{3} mgR _{ e }$
C$\frac{2}{3} mgR _{ e }$
D$\frac{1}{2} mgR _{ e }$
(JEE MAIN-2023)
Solution
$U =\frac{- GM _{ e }^{ m }}{ r }$
$U _{ i } =\frac{-G M_e m }{ R _{ e }}$
$U _{ f } =\frac{- GM _{ e } m }{\left( R _{ e }+ h \right)}=\frac{-G M_e m }{ R _{ e }+2 R _{ e }}$
$\frac{- GM _{ e } m }{3 R _{ e }}$
Increase in internal energy $\Delta U = U _{ f }- U _{ i }$
$=\frac{2}{3} \frac{ GM _{ e } m }{ R _{ e }}$
$\frac{2}{3} \frac{ GM _{ e }}{ R _{ e }^2} mR _{ e }$
$=\frac{2}{3} mgR _{ e }$
$U _{ i } =\frac{-G M_e m }{ R _{ e }}$
$U _{ f } =\frac{- GM _{ e } m }{\left( R _{ e }+ h \right)}=\frac{-G M_e m }{ R _{ e }+2 R _{ e }}$
$\frac{- GM _{ e } m }{3 R _{ e }}$
Increase in internal energy $\Delta U = U _{ f }- U _{ i }$
$=\frac{2}{3} \frac{ GM _{ e } m }{ R _{ e }}$
$\frac{2}{3} \frac{ GM _{ e }}{ R _{ e }^2} mR _{ e }$
$=\frac{2}{3} mgR _{ e }$
Standard 11
Physics
Similar Questions
Match the column $-I$ with column $-II$ For a satellite in circular orbit,
| Column $-I$ | Column $-II$ |
| $(A)$ Kinetic energy | $(p)$ $ – \frac{{G{M_E}m}}{{2r}}$ |
| $(B)$ Potential energy | $(q)$ $\sqrt {\frac{{G{M_E}}}{r}} $ |
| $(C)$ Total energy | $(r)$ $ – \frac{{G{M_E}m}}{{r}}$ |
| $(D)$ Orbital energy | $(s)$ $ \frac{{G{M_E}m}}{{2r}}$ |
(where $M_E$ is the mass of the earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
hard
