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Gravitation — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsGravitation4 Marks
Question
Using expression for change in potential energy, show that gravitational potential energy of the system of object of mass $m$ and the Earth with separation of $r$ is, $-\frac{ GMm }{r}$
✓

Answer

$1.$ Change in $P.E.$ for a system of Earth and mass is given by,
$\Delta U=\operatorname{GMm}\left(\frac{1}{r_i}-\frac{1}{r_f}\right)$
$2.$ For gravitational force, point of zero potential energy is taken to be at $r = \infty .$
$3.$ Hence, $U(r_i) = 0$ at $r_i=\infty$
Final point $r_f $is the point where the potential energy of the system is to be determined.
$4. \ At r_f = r$
$ U ( r )_{\text {grav. }} = GMm \left(\frac{1}{ r _{ i }}-\frac{1}{ r _{ f }}\right)$
$ = GMm \left(\frac{1}{\infty}-\frac{1}{ r }\right)$
$=-\frac{G M m}{r}$
This is gravitational potential energy of the system of object of mass $m$ and Earth of mass $M$ having separation $r$ $($between their centres of mass$).$

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