Work Done Against Gravity
Category : JEE Main & Advanced
If the body of mass \[m\] is moved from the surface of earth to a point at distance \[h\] above the surface of earth, then change in potential energy or work done against gravity will be
\[T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \,\,\sqrt{\frac{{{r}^{3}}}{g{{R}^{2}}}}\]
\[\Rightarrow \] \[W=GMm\left[ \frac{1}{R}-\frac{1}{R+h} \right]\] [As \[{{r}_{1}}=R\] and \[v=\sqrt{\frac{GM}{r}}\]]
\[\Rightarrow \] \[W=\frac{GMmh}{{{R}^{2}}\left( 1+\frac{h}{R} \right)}\]\[=\frac{mgh}{1+\frac{h}{R}}\] [As \[\frac{GM}{{{R}^{2}}}=g\]]
(i) When the distance \[h\] is not negligible and is comparable to radius of the earth, then we will use above formula.
(ii) If \[h=nR\] then \[W=mgR\left( \frac{n}{n+1} \right)\]
(iii) If \[h=R\] then \[W=\frac{1}{2}mgR\]
(iv) If \[h\] is very small as compared to radius of the earth then term \[h/R\] can be neglected
From \[W=\frac{mgh}{1+h/R}\]\[=mgh\] \[\left[ \text{As }\,\,\frac{h}{R}\to 0 \right]\]
You need to login to perform this action.
You will be redirected in
3 sec