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Answer with Explanation:
Let rest mass [tex]m_0[/tex] at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.
Mass of moon=m
Distance between the center of moon and center of planet=D
Mass of planet=M
We are given that net force on an object will be zero
a.We have to derive an expression for x in terms of m, M and D.
We know that gravitational force=[tex]\frac{GmM}{r^2}[/tex]
Distance of P from moon=D-x
[tex]F_m[/tex]=Force applied on rest mass due to m
[tex]F_m[/tex]=Force on rest mass due to mas M
[tex]F_M=F_m[/tex] because net force is equal to 0.
[tex]F_m=F_M[/tex]
[tex]\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}[/tex]
[tex]\frac{m}{(D-x)^2}=\frac{M}{x^2}[/tex]
[tex]\frac{x^2}{(D-x)^2}=\frac{M}{m}[/tex]
[tex]\frac{x}{D-x}=\sqrt{\frac{M}{m}}[/tex]
Let [tex]R=\sqrt{\frac{M}{m}}[/tex]
Then, [tex]\frac{x}{D-x}=R[/tex]
[tex]x=DR-xR[/tex]
[tex]x+xR=DR[/tex]
[tex]x(1+R)=DR[/tex]
[tex]x=\frac{DR}{1+R}[/tex]
b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=[tex]\frac{2}{3}D[/tex]
Net force is zero
[tex]F_m=F_M[/tex]
[tex]\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}[/tex]
[tex]\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}[/tex]
[tex]\frac{M}{m}=4[/tex]
Hence, the ratio R of the mass of the planet to the mass of the moon=4:1
A) Th expression for x in terms of m, M and D is;
x = [tex]\frac{D\sqrt{m}}{\sqrt{m} + \sqrt{M}}[/tex]
B) The ratio R of the mass of the planet to the mass of the moon is;
R = M/m = 4
We are given;
Mass of moon = m
Mass of planet = M
Center to center distance from moon to planet = D
Net force is zero at a distance x from the planet's center.
- A) We know that formula for Gravitational Force is;
F_g = GMm/r²
Now, since x is a distance from the center of planet, then its' distance from
the moon is; D - x
This means;
For the moon; F_g,moon = Gm₁m/(D - x)²
For the planet; F_g,planet = Gm₁M/x²
Where m₁ is mass of the earth
We are told that net force on the object is zero.
Thus; F_g,moon = F_g,planet
⇒ Gm₁m/(D - x)² = Gm₁M/x²
G and m₁ will cancel out to give us;
m/(D - x)² = M/x²
Cross multiply to get;
mx² = M(D - x)²
Taking square root of both sides gives;
(√m)x = (√M) × (D - x)
Expand the right side to get;
(√m)x = D(√M) - x(√M)
(√m)x + x(√M) = D(√M)
x((√m) + (√M)) = D(√M)
x = [tex]\frac{D\sqrt{m}}{\sqrt{m} + \sqrt{M}}[/tex]
B) Net force is at ⅔D from the planet. Thus, x = ⅔D from the planet.
From earlier, we saw that; mx² = M(D - x)². Thus;
M/m = x²/(D - x)²
Putting ⅔D for x gives;
M/m = (⅔D)²/(D - ⅔D)²
M/m = ⁴/₉(D²)/(¹/₉D²)
This will simplify out to give;
M/m = 4
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