Question

The gravitational force of attraction between two bodies at a certain distance is 70 N. If the distance of separation between them is doubled, then find the percentage change in force of attraction
(1) Decreases by 25 %
(2) Decreases by 75%
(3) Increases by 50%
14) Increases by 75%​

Answer 1

Answer:

Given:

Gravitational force between 2 bodies = 70 N

Now , distance of separation is doubled.

To find:

% change in the Gravitational force

Formulas used:

Gravitational force = (G × m1 × m2)/d²,

where m1 and m2 are the masses of the bodies , G is the universal gravitational constant.

and d is the separation distance.

Calculation:

Initially,

F = (G × m1 × m2)/d²

=> 70 = (G × m1 × m2) / d²

In the 2nd case,

d" = 2d .............(as per question)

So,

F" = (G × m1 × m2)/(d")²

=> F" = (G × m1 × m2)/(2d)²

=> F" = ¼ [(G × m1 × m2)/d²]

=> F" = ¼ × 70

=> F" = 17.5 N

So decrease in force = 70 - (17.5)

=> ∆F = 52.5 N

% decrease = ∆F/F × 100%

=> % decrease = 52.5/70 × 100%

=> % decrease = 75 %

So correct option is 2) decreases by 75%

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

The gravitational force of attraction between two bodies at a certain distance is 70 N. If the distance of separation between them is doubled, then find the percentage change in force of attraction?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial Force (F₁) = 70 N.
• Let the separation between the masses be "d₁".
• Distance is Doubled, I.e. d₂ = 2d₁
• Let the Force obtained after Doubling the distance be F₂.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From Universal Law of Gravitation,

$\large{\boxed{\tt F = \dfrac{GMm}{r^2}}}$

Substituting the values,

$\large{\tt \leadsto F_1 = \dfrac{GMm}{d_1^2}}$

$\large{\tt \leadsto F_1 = \dfrac{GMm}{d_1^2}}$

$\large{\leadsto{\underline{\underline{\tt F_1 = \dfrac{GMm}{d_1^2}}}} \: \tt -----(1)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, The Distance is Doubled,

⇒ d₂ = 2d₁

From Universal Law of Gravitation,

$\large{\boxed{\tt F = \dfrac{GMm}{r^2}}}$

Substituting the values,

$\large{\tt \leadsto F_2 = \dfrac{GMm}{d_2^2}}$

$\large{\tt \leadsto F_2 = \dfrac{GMm}{(2 d_1)^2}}$

$\large{\tt \leadsto F_2 = \dfrac{GMm}{4 .d_1^2}}$

$\large{\tt \leadsto F_2 = \dfrac{1}{4} \times \dfrac{GMm}{d_1^2}}$

Comparing with Equation (1) we get.

$\large{\tt \leadsto F_2 = \dfrac{1}{4} \times F_1}$

$\large{\tt \leadsto F_1 = 4 \times F_2}$

$\large{\boxed{\tt F_1 = 4F_2}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Finding the Percentage change,

$\large{\boxed{\tt \Delta F \% = \dfrac{F_2 - F_1}{F_1} \times 100}}$

Substituting the values,

$\large{\tt \leadsto \Delta F \% = \dfrac{F_2 - 4F_2}{4F_2} \times 100}$

$\large{\tt \leadsto \Delta F \% = \dfrac{- 3F_2}{4F_2} \times 100}$

$\large{\tt \leadsto \Delta F \% = \dfrac{ - 3 \cancel{F_2}}{4\cancel{F_2}} \times 100}$

$\large{\tt \leadsto \Delta F \% = \dfrac{- 3}{4} \times 100}$

$\large{\tt \leadsto \Delta F \% = \dfrac{- 3}{\cancel{4}} \times \cancel{100}}$

$\large{\tt \leadsto \Delta F \% = - 3 \times 25}$

$\huge{\boxed{\boxed{\tt \Delta F\% = - 75 \% }}}$

Note:-

• Here negative sign indicates that the Force is Decreasing.

So, the percentage change in force of attraction is Decreases by 75% (Option- 2)

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