If the distance between the sun and the earth is increased by three times then attraction between two will

Question

If the distance between the sun and the earth is increased by three times then attraction between two will (A) remain constant (B) decrease by 63% (C) increase by 63% (D) decrease by 89%

Tardigrade Admin · Accepted Answer

F=(G m1 m2/r2) and F'=(G m1 m2/(3 r)2)=(F/9) ∴ % decrease in F=((F-F'/F)) × 100 =(8/9) × 100 =89 % Thus, attraction force between sun and earth is decreased by 89 %.

Q. If the distance between the sun and the earth is increased by three times then attraction between two will

remain constant

decrease by 63%

increase by 63%

decrease by 89%

$F = \frac{G m _{1} m _{2}}{r ^{2}}$
and $F^{'} = \frac{G m _{1} m _{2}}{( 3 r ) ^{2}} = \frac{F}{9}$
$∴ %$ decrease in $F = (\frac{F - F ^{'}}{F}) \times 100$
$= \frac{8}{9} \times 100$
$= 89%$
Thus, attraction force between sun and earth is decreased by $89%$ .

Q. If the distance between the sun and the earth is increased by three times then attraction between two will

remain constant

decrease by 63%

increase by 63%

decrease by 89%

F=r2Gm1​m2​​ and F′=(3r)2Gm1​m2​​=9F​ ∴% decrease in F=(FF−F′​)×100 =98​×100 =89% Thus, attraction force between sun and earth is decreased by 89%.

$F = \frac{G m _{1} m _{2}}{r ^{2}}$
and $F^{'} = \frac{G m _{1} m _{2}}{( 3 r ) ^{2}} = \frac{F}{9}$
$∴ %$ decrease in $F = (\frac{F - F ^{'}}{F}) \times 100$
$= \frac{8}{9} \times 100$
$= 89%$
Thus, attraction force between sun and earth is decreased by $89%$ .