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(A) The gravitational force exerted by a body of mass $M$ on a mass $m$ at a distance $r$ is $F = \frac{GMm}{r^2}$.The difference in force $\Delta F$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The gravitational force exerted by a body of mass $M$ on a mass $m$ at a distance $r$ is $F = \frac{GMm}{r^2}$.The difference in force $\Delta F$ between the nearest point $(r-R_e)$ and the farthest point $(r+R_e)$ on Earth is given by $\Delta F = \frac{GMm}{(r-R_e)^2} - \frac{GMm}{(r+R_e)^2} \approx \frac{2GMm}{r^3} (2R_e) = \frac{4GMmR_e}{r^3}$,where $R_e$ is the radius of the Earth.For the moon: $\Delta F_1 = \frac{4GM_m m R_e}{r_1^3}$.For the sun: $\Delta F_2 = \frac{4GM_s m R_e}{r_2^3}$.Taking the ratio:$\frac{\Delta F_1}{\Delta F_2} = \frac{M_m}{M_s} 	imes \left( \frac{r_2}{r_1} ight)^3$.Given values: $M_m = 8 	imes 10^{22} \, kg$,$M_s = 2 	imes 10^{30} \, kg$,$r_1 = 0.4 	imes 10^6 \, km$,$r_2 = 150 	imes 10^6 \, km$.$\frac{\Delta F_1}{\Delta F_2} = \left( \frac{8 	imes 10^{22}}{2 	imes 10^{30}} ight) 	imes \left( \frac{150 	imes 10^6}{0.4 	imes 10^6} ight)^3 = (4 	imes 10^{-8}) 	imes (375)^3$.$(375)^3 = 52,734,375$.$\frac{\Delta F_1}{\Delta F_2} = 4 	imes 10^{-8} 	imes 5.27 	imes 10^7 \approx 2.1$.Thus,the closest integer is $2$.

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	$(s)$ $\frac{GM_Em}{2r}$

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

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