Let the field be zero at a point at distance x from M₁.
\(\therefore \frac{GM_1}{x^2}=\frac{GM_2}{(d-x)^2}\)
\(\therefore \frac{x}{d-x}=\sqrt {\frac{M_1}{M_2}}\Rightarrow x\sqrt {M_2}=\sqrt {M_1}.d-x\sqrt {M_1}\)
\(x[\sqrt {M_!}+\sqrt {M_2}]=\sqrt {M_1}.d\)
\(x=\frac{d\sqrt {M_2}}{\sqrt {M_1}+\sqrt {M_2}}\)
\(d-x=\frac{d\sqrt {M_2}}{\sqrt {M_1}+\sqrt {M_2}}\)
Potential at this point due to both the masses will be
\(=-\frac{GM_1}{x}-\frac{GM_2}{(d-x)}=-G[\frac{M_1(\sqrt {M_1}+\sqrt {M_2})}{d\sqrt {M_1}}+\frac{M_2(\sqrt {M_1}+\sqrt {M_2})}{d\sqrt {M_2}}]\)
\(=-\frac{G}{d}(\sqrt {M_1}+\sqrt {M_2})^2=\frac{G}{d}(M_1+M_2+2\sqrt {M_1}\sqrt {M_2}).\)

Mass of Sun, M = 2 x 10³⁰ kg; Mass of Earth, m = 6 x 10²⁴ kg
Distance between Sun and Earth, r = 1.5 x 10¹¹ m

Let at the point P, the gravitational force on the rocket due to Earth 
= gravitational force on the rocket due to Sun
Let x = distance of the point P from the Earth
Then \(\frac{G_m}{x^2}=\frac{GM}{(r-x)^2}\)
\(\Rightarrow\frac{(r-x)^2}{x^2}=\frac{M}{m}=\frac{2\times10^{30}}{6\times10^{24}}=\frac{10^6}{3}\)
\(\frac{r-x}{x}=\frac{10^3}{\sqrt 3}\Rightarrow \frac{r}{x}=\frac{10^3}{\sqrt 3}+1\simeq\frac{10^3}{\sqrt 3}\)
\(x=\frac{\sqrt 3 r}{10^3}=\frac{1.732\times1.5\times10^{11}}{10^3}=2.6\times10^8m.\)

For a satellite of jupiter, orbital period,T₁= 1.769 days = 1.769 x 24 x 60 x 60s
Radius of the orbit of satellite, r₁ = 4.22 x 10⁸m
Mass of Jupiter, M₁ is given by M₁ = \(\frac{4\pi^2r_1^3}{GT_1^2}=\frac{4\pi^2\times(4.22\times10^8)}{G\times(1.769\times24\times60\times60)^2}\) ...(1)
We know that the orbital period of earth around the sun, T=1 year=365.25 x 24 x 60 x 60s
orbital radius,r = 1 A.U.= 1.496 x 10¹¹m
Mass of sun is given by \(M=\frac{4\pi^2r^3}{GT^2}=\frac{4\pi^2\times(1.496\times10^{11})^3}{G\times(365.25\times24\times60\times60)^2}\) ....(2)
Dividing eqn. (ii) by (i), we get
\(\frac{M}{M_1}=\frac{4\pi^2\times(1.496\times10^{11})^3}{G\times(365.25\times24\times60\times60)^2}\times\frac{G\times(1.769\times24\times60\times60)^2}{4\pi^2\times(4.22\times10^8)}=1046\)
or \(\frac{M}{M_1}=\frac{1}{1046}=\frac{1}{1000}\Rightarrow m_1=\frac{1}{1000}M.\) 

40 kg

\(3.468\times 10^{ -10 }N\)

As discussed in the hint section, Kepler’s law states that
\([{{T}^{2}}=k\times {{a}^{3}}]\) where [T] is the period of revolution of a planet, [a] is the semi-major axis of the planet and [K] is a constant of proportionality
Upon researching, we found that
The semi-major axis of the earth \([\left( {{a}_{earth}} \right)=149.6\times {{10}^{6}}km]\). Similarly, the semi-major axis of Neptune \([\left( {{a}_{neptune}} \right)=4495.06\times {{10}^{6}}km]\)
The period of revolution of the earth around the sun is one year. Substituting these values, we get the expression of Kepler’s Law for both planets as follows
\({{\left( {{T}_{earth}} \right)}^{2}}=k\times {{\left( {{a}_{earth}} \right)}^{3}}\) ...(1)
\( {{\left( {{T}_{neptune}} \right)}^{2}}=k\times {{\left( {{a}_{neptune}} \right)}^{3}}\) ....(2)
Dividing the two equations, we get
\([{{\left( \dfrac{{{T}_{neptune}}}{{{T}_{earth}}} \right)}^{2}}={{\left( \dfrac{{{a}_{neptune}}}{{{a}_{earth}}} \right)}^{3}}]\)
Substituting the values of the semi-major axis of the planets and the period of revolution of the earth, we get
\( {{\left( \dfrac{{{T}_{neptune}}}{1year} \right)}^{2}}={{\left( \dfrac{4495.06\times {{10}^{6}}km}{149.6\times {{10}^{6}}km} \right)}^{3}} \)
\( \Rightarrow {{\left( \dfrac{{{T}_{neptune}}}{1} \right)}^{2}}={{(30.05)}^{3}} \)
\( \Rightarrow {{T}_{neptune}}={{(30.05)}^{\frac{3}{2}}}=164.72years \)

(i)\({ R }_{ 2 }\approx 30{ R }_{ 1 }\)
(ii)\({ R }_{ 2 }\approx 30{ R }_{ 1 }\)

From law of gravitation, \(F=\frac { GMm }{ { R }^{ 2 } } \)
Where, m = mass of object on surface of the earth.
R = radius of the earth.
M = mass of the earth.
Force experienced by object, F = mg
Using the two forces, we have
\(mg=\frac { GMm }{ { R }^{ 2 } } \quad or\quad M=\frac { gR^{ 2 } }{ G } \)

Height of the earth's satellite\(h=\left( \frac { { T }^{ 2 }R^{ 2 }g }{ 4\pi ^{ 2 } } \right) -R\)
Given,T= 90  min = \(90\times 60=5400s,R=6380km\)
\(g=9.8m/{ s }^{ 2 }=9.8\times { 10 }^{ -3 }km/{ s }^{ 2 }\)
Thus
 \(h=\left[ \frac { (5400)^{ 2 }\times (6380)^{ 2 }\times (9.8\times { 10 }^{ -3 } }{ 4\times 9.87 } \right] ^{ 1/3 }km-6380\quad km\)
or h = (6655 - 6380)km = 275 km

We are given that
Mass of the earth, M = \(6.0\times { 10 }^{ 24 }\)kg
Radius of the earth, R = 6400 km
Height of the satellite from the earth's surface.
h = 36000 km
Distance of the satellite from the centre of the earth.
r = R + h = 6400km + 36000km
r = 42400km = \(4.24\times { 10 }^{ 7 }\)m
If V is the potential at the site of the satellite.
\(V=-\frac { GM }{ r } =-5.12\times { 10 }^{ 7\ }J/kg\)
\(\\ V=-9.4\times { 10 }^{ 6 }J/kg\)

Depth of sea, d =7 km, g = 9.8 ms^-2
Radius of the earth, \(R=\frac { D }{ 2 } =\frac { 12800 }{ 2 } km=6400km\)
Value of g at bottom of sea
\({ g }_{ d }=g\left( 1-\frac { d }{ R } \right) =9.8(1-\frac { 7 }{ 6400 } ){ ms }^{ -2 }\)
\(\\ { g }_{ d }=\frac { 9.8\times 6393 }{ 6400 } { ms }^{ -2 }=9.789{ ms }^{ -2 }\)
Note: Acceleration due to gravity also vary due to the rotation of earth, if a body of mass m lying at a point whose latitude is \(\lambda \) , then rotation of earth(angular speed \(\omega \) ), the apparent acceleration due to gravity on body is given by \({ g }^{ ' }=g-{ \omega }^{ 2 }R{ cos }^{ 2 }\lambda \)