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(B) Using the Work-Energy Theorem: $\Sigma W = \Delta K$.Work done by the gravitational force $F = -G \frac{Mm}{r^{2.1}}$ as the satellite moves from

Vedclass · Accepted Answer

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

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(B) Using the Work-Energy Theorem: $\Sigma W = \Delta K$.Work done by the gravitational force $F = -G \frac{Mm}{r^{2.1}}$ as the satellite moves from $r = a$ to $r = b$ is:$W = \int_{a}^{b} F \cdot dr = \int_{a}^{b} -\frac{GMm}{r^{2.1}} dr$.$W = -GMm \int_{a}^{b} r^{-2.1} dr = -GMm \left[ \frac{r^{-1.1}}{-1.1} \right]_{a}^{b}$.$W = \frac{GMm}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right]$.According to the Work-Energy Theorem,$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.Equating the two expressions:$\frac{GMm}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right] = \frac{1}{2}m(v^2 - u^2)$.Dividing by $m/2$:$v^2 - u^2 = \frac{2GM}{1.1} \left[ \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right]$.$v = \sqrt{u^2 + \frac{2}{1.1}GM \left( \frac{1}{b^{1.1}} - \frac{1}{a^{1.1}} \right)}$.

$A$ satellite of mass $m$ is at a distance of $a$ 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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