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(C) The potential energy $U(r)$ is given by $U(r) = -\int_{\infty}^{r} F dr = -\int_{\infty}^{r} -\frac{GMm}{r^{2.1}} dr = GMm \int_{\infty}^{r} r^{-2

Vedclass · Accepted Answer

$\sqrt{u^2 + \frac{2}{1.1} GM \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)}$

Vedclass · Answer

(C) The potential energy $U(r)$ is given by $U(r) = -\int_{\infty}^{r} F dr = -\int_{\infty}^{r} -\frac{GMm}{r^{2.1}} dr = GMm \int_{\infty}^{r} r^{-2.1} dr$.Evaluating the integral: $U(r) = GMm \left[ \frac{r^{-1.1}}{-1.1} \right]_{\infty}^{r} = -\frac{GMm}{1.1 r^{1.1}}$.By the law of conservation of energy: $K_i + U_i = K_f + U_f$.$\frac{1}{2}mu^2 - \frac{GMm}{1.1 a^{1.1}} = \frac{1}{2}mv^2 - \frac{GMm}{1.1 b^{1.1}}$.Dividing by $m/2$: $u^2 - \frac{2GM}{1.1 a^{1.1}} = v^2 - \frac{2GM}{1.1 b^{1.1}}$.Solving for $v$: $v^2 = u^2 + \frac{2GM}{1.1} \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)$.Thus,$v = \sqrt{u^2 + \frac{2}{1.1} GM \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)}$.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

$A$ satellite of mass $m$ is at a distance $a$ from a star of mass $M$. The speed of the satellite is $u$. Suppose the law of universal gravity is $F = -G \frac{Mm}{r^{2.1}}$ instead of $F = -G \frac{Mm}{r^2}$. Find the speed of the satellite when it is at a distance $b$ from the star.

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