Physics, asked by ad200421, 11 months ago

Gravitational force between two point masses m1 and m2 placed at a distance r is given by
F = Gm1m2/r^2
; where G is an universal constant. An object of mass M is divided into two parts, which are placed at distance r. Find the mass of both parts if gravitation force of attraction is maximum between them.

Answers

Answered by sharansuryas2s

25

Answer:

m1 = m2 = M/2

Explanation:

M is divided into 2 parts. M = m1 + m2

The force between the two parts will be :

F = Gm1m2/r²

F = Gm1(M-m1)/r²

F = GMm1/r² - Gm1²/r²

For Force to be maximum, dF/dm1 = 0

dF/dm1 = GM/r² -2Gm1/r² = 0

(M - 2m1) = 0

m1 = M/2

Hence the masses are m1 = M/2, m2 = M/2

Answered by brainlyaryan12

46

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✪ $\huge{\red{\underline{\overline{\mathbf{Question}}}}}$ ✪

→Gravitational force between two point masses m1 and m2 placed at a distance r is given by

F = Gm1m2/r^2

; where G is an universal constant. An object of mass M is divided into two parts, which are placed at distance r. Find the mass of both parts if gravitation force of attraction is maximum between them.

✪ $\huge{\green{\underline{\overline{\mathbf{Answer}}}}}$ ✪

⇒Given:

⇒G and r are Constant
⇒M is the Combined mass

⇒To Find:

⇒Masses of the Two parts

⇒Solution:-

⇒Let First Mass $m_1$ = m

⇒ $m_2$ = (M-m)

Now:-

$\Large{F=\frac{Gm_1m_2}{r^2}}$

⇒ $\frac{GM}{r^2}=\frac{Gm(M-m)}{r^2}$

⇒ $\frac{GM}{r^2}=\frac{GMm}{r^2}-\frac{Gm^2}{r^2}$

On Differentiation of Both Sides:

⇒ $\frac{d(GM)}{dm(r^2)}=\frac{d(GMm)}{dm(r^2)}-\frac{d(Gm^2)}{dm(r^2)}$

⇒ $\cancel{\frac{G}{r^2}}\bigg[\frac{d(M)}{dm}\bigg]=\cancel{\frac{G}{r^2}}\bigg[ \frac{d(Mm)}{dm}\bigg]-\cancel{\frac{G}{r^2}}\bigg[ \frac{d(m^2)}{dm</p><p>}\bigg]$

⇒ $\large{0=M-2m}$

⇒ $\huge{\pink{\overbrace{\underbrace{\red{m=\small{\frac{M}{2}}}}}}}$

≿━━━━━━━━━༺❀༻━━━━━━━━━≾

Formulas Used :-

⇒ $\large{\frac{d(k)}{dx}=0}$ [k=Constant]
⇒ $\large{\frac{d(x)}{dx}=1}$
⇒ $\large{\frac{d(x^n)}{dx}=nx^{n-1}}$

≿━━━━━━━━━༺❀༻━━━━━━━━━≾

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