Question

Answer quicklyyyyyyyyyyyyyyyyyy

Answer 1

Answer:

$F_g = \frac{GM_1M_2}{R^2}$

Masses increased by 25% ---> M becomes 5M/4

Distance reduced by 20% ---> R becomes 4R/5

$F'_g = \frac{G*\frac{5}{4} M_1*\frac{5}{4} M_2}{(\frac{4R}{5} )^2} \\\\F'_g = \frac{\frac{25}{16} GM_1M_2}{\frac{16R^2}{25} } \\\\F'_g = \frac{625}{256} \frac{GM_1M_2}{R^2} } \\\\F'_g = \frac{625}{256} F_g$

Change in the force is

ΔF = $F'_g - F_g = (\frac{625}{256} - 1) \frac{GM_1M_2}{R^2} = (\frac{625-256}{256}) \frac{GM_1M_2}{R^2}\\\\= (\frac{369}{256}) \frac{GM_1M_2}{R^2} = 1.44 \frac{GM_1M_2}{R^2}$

We can see that the change in force is nearly increased by 44%