Question

when the distance between two bodies is doubled and mass of each body is also doubled then the gravitational force will be
A)doubled
B)remained constant
C)trebled
D)quadrupled


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Answer 1

Answer:

• The gravitational force (F) will Same/Remain constant.

Assumption:

Case-1

1. Let the mass of one body be " m₁ ".
2. Let the mass of second body be " m₂ "
3. Let the Distance be " r₁ "
4. Let the Force be " F₁ "

Case-2

1. Let the mass of one body  be " m₃ ".
2. Let the mass of second body be " m₄ "
3. Let the Distance be " r₂ "
4. Let the Force be " F₂ "

Explanation:

$\rule{300}{1.5}$

Case-1

From the formula we know,

$\large \bigstar\; \boxed{\tt F_1=\dfrac{G\;m_1\;m_2}{(r_1)^2}}$

$\frak{Here}\begin{cases}\sf{m_1}\text{ Denotes Mass of 1st Body}\\ \sf{m_2}\text{ Denotes Mass of 2nd Body}\\\sf{r_1}\text{ Denotes Distance}\end{cases}$

Now,

$\displaystyle\dashrightarrow\tt F_1 = \dfrac{G\;m_1\;m_2}{(r_1)^2} \quad\dfrac{\quad}{}[1]$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, the formula becomes,

$\large \bigstar\; \boxed{\tt F_2=\dfrac{G\;m_3\;m_4}{(r_2)^2}}$

$\frak{Here}\begin{cases}\sf{m_3}\text{ Denotes changed Mass of 1st Body}\\ \sf{m_2}\text{ Denotes Changed Mass of 2nd Body}\\\text{ \tt r_2 Denotes Distance}\end{cases}$

Now,

$\displaystyle\dashrightarrow\tt F_2 = \dfrac{G\;m_3\;m_4}{(r_2)^2}$

Now, we Know, (From the Question)

• m₃ = 2 m₁
• m₄ = 2 m₂
• r₂ = 2 r₁

Substituting in the above equation we get,

$\displaystyle\dashrightarrow\tt F_2 = \dfrac{G\times 2\;m_1\times 2\;m_2}{(2\;r_1)^2}\\\\\\\dashrightarrow\tt F_2 = \dfrac{4\; (G\;m_1\;m_2)}{4\; (r_1^{2})}\\\\\\\dashrightarrow\tt F_2 = \dfrac{4}{4}\times \dfrac{G\;m_1\;m_2}{r_1^2}\\\\\\\dashrightarrow\tt F_2 = \dfrac{G\;m_1\;m_2}{r_1^2}\\\\\\\dag\quad \blue{\tt \dfrac{G\;m_1\;m_2}{r_1^2}= F_1} \;\;[Equation-1] \\\\\\\dashrightarrow\tt F_2 = F_1\\\\\\\dashrightarrow \large {\underline{\boxed{\red{\tt F_2 = F_1}}}}$

∴ The gravitational force (F) will Same/Remain constant.

$\rule{300}{1.5}$

Answer 2

Answer:

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