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(C) The orbital radius $r$ is given by $r = R_{planet} + 	ext{altitude}$. Let the planet radius be $R_p$. However,usually in such problems,altitude i

Vedclass · Accepted Answer

Ratio of total energy of $A$ and $B$ is $3 : 2$

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(C) The orbital radius $r$ is given by $r = R_{planet} + 	ext{altitude}$. Let the planet radius be $R_p$. However,usually in such problems,altitude is given from the surface. If altitude is $h$,then $r = R_p + h$. Assuming the given values $3R$ and $5R$ are the distances from the center of the planet ($r_A = 3R$ and $r_B = 5R$): Potential Energy $U = -\frac{GMm}{r}$ Kinetic Energy $K = \frac{GMm}{2r}$ Total Energy $T = -\frac{GMm}{2r}$ For satellite $A$ at $r_A = 3R$: $T_A = -\frac{GMm}{6R}$ For satellite $B$ at $r_B = 5R$: $T_B = -\frac{GMm}{10R}$ Ratio $\frac{T_A}{T_B} = \frac{10R}{6R} = \frac{5}{3}$ Checking Kinetic Energy: $\frac{K_A}{K_B} = \frac{r_B}{r_A} = \frac{5R}{3R} = \frac{5}{3}$ Checking Potential Energy: $\frac{U_A}{U_B} = \frac{r_B}{r_A} = \frac{5R}{3R} = \frac{5}{3}$ Given the options provided and the standard interpretation,if $r_A = 4R$ and $r_B = 6R$ (altitude $3R$ and $5R$ from surface with planet radius $R$): $r_A = 4R, r_B = 6R$ $\frac{T_A}{T_B} = \frac{r_B}{r_A} = \frac{6R}{4R} = \frac{3}{2}$. Thus,option $C$ is correct.

Two satellites $A$ and $B$ of same mass are moving around a planet at an altitude of $3R$ and $5R$ respectively. Select the correct statement.

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