Physics - Gravitation Question with Solution | TestHub

$$\begin{array}{rlr}{\mathrm{w}}_1& =\frac{(\mathrm{GM}{)}^{1/2}}{(\mathrm{r}-\Delta \mathrm{r}{)}^{3/2}}{\mathrm{w}}_2& =\frac{(\mathrm{GM}{)}^{1/2}}{{\mathrm{r}}^{3/2}}\end{array}$$

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$$\begin{array}{rlr}& {\mathrm{r}}_2=\mathrm{r}& {\mathrm{r}}_1=\mathrm{r}-\Delta \mathrm{r}\end{array}$$

Angular velocity of approach $={\mathrm{w}}_1-{\mathrm{w}}_2$

$$\begin{array}{rlrl}& =(\mathrm{GM}{)}^{1/2}\left[\frac{r^{3/2}-(r-\Delta r{)}^{3/2}}{r^{3/2}(r-\Delta r{)}^{3/2}}\right]& =(\mathrm{GM}{)}^{1/2}\times \frac{\left[1-\left(1-\frac{3\Delta r}{2r}\right)\right]}{r^{3/2}{\left(1-\frac{\Delta r}{r}\right)}^{3/2}}& =\frac{3(\mathrm{GM}{)}^{1/2}\Delta r}{2r^{5/2}}\end{array}$$

Time period of approach $=\frac{2\pi }{{\mathrm{w}}_1-{\mathrm{w}}_2}=\frac{4\pi {\mathrm{r}}^{5/2}}{3(\mathrm{GM}{)}^{1/2}\Delta \mathrm{r}}$