d = 4 cm

−6.25 × 10⁷J/kg

5 kms\(^{ -1 }\)

6.67 × 10⁻⁹N

Gravitational potential energy, i.e
\(U_{ h }=-\frac { GMm }{ R+h } =-\frac { gR^{ 2 }m }{ R+h } \left[ where,g=\frac { GM }{ R^{ 2 } } \right] \)
\(=-\frac { gR^{ 2 }m }{ R\left( 1+\frac { h }{ R } \right) } =-\frac { mgR }{ 1+\frac { h }{ R } } \)

The value of g is more at poles than at the equator. Therefore, a body weighs more at poles than at equator.

At the mountain, g decreases and time period of the pendulum clock increases at T=2π\(\sqrt{l/g} \). On the other hand, a spring in the wrist watch remains unaffected by the variation of g.

The value of g in mine is less than that on the surface of the earth.

The value of escape velocity on the moon is small as compared to the value of the earth only 2.5 kms ^-1 So,thw molecules of air escape easily from the surface of the moon,hence there is no atmosphere.

\(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } }\)
\( \\ =6.67\times 10^{ -8 })\left( \frac { 1\times 1 }{ 1^{ 2 } } \right) dyne\)
\(\\ =6.67\times 10^{ -8 }\quad dyne=\frac { 6.67\times 10^{ -8 } }{ 980 } \)
\(=7\times 10^{ -11 }g-wt\)

1.972 × 10³⁰kg

Tides arise in the oceans due to the force of attraction between the moon and sea water.

\(T^{ 2 }\alpha R^{ 3 }\)
Hence, \(\frac { { T }_{ 1 }^{ 2 } }{ { T }_{ 2 }^{ 2 } } =\frac { { R }_{ 1 }^{ 3 } }{ { R }_{ 2 }^{ 3 } } \Rightarrow { T }_{ 2 }^{ 2 }=\left[ \frac { { R }_{ 2 } }{ { R }_{ 1 } } \right] ^{ 3 }T_{ 1 }^{ 2 }\)
\(\Rightarrow T_{ 2 }=T_{ 1 }\left[ \frac { R_{ 2 } }{ { R }_{ 1 } } \right] ^{ 3/2 }=365\left( \frac { R/2 }{ R } \right) ^{ 3/2 } \)
\(=365\times \frac { 1 }{ \sqrt [ 2 ]{ 2 } } =129\ days\)

According to Kepler's third law of planetary motion,
\({ T }^{ 2 }\ \alpha \ { r }^{ 3 }\)
\(\\ Thus,\ \frac { { T }_{ S }^{ 2 } }{ { T }_{ E }^{ 2 } } =\frac { { r }_{ S }^{ 3 } }{ { r }_{ E }^{ 3 } } \ or \left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2 }=\left( \frac { { r }_{ S } }{ { r }_{ E } } \right) ^{ 3 }\)
\(\\ \left( \frac { { r }_{ S } }{ { r }_{ E } } \right) =\left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2/3 }\\\)
\( \\ { r }_{ S }=\left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2/3 }\times { r }_{ E }\)
\(\\ As,\quad \left( \frac { { T }_{ S } }{ { T }_{ E } } \right) =29.5\)
\(\\ { r }_{ E }=1.5\times 10^{ 8 }km\)
\(\\ { r }_{ S }=(29.5)^{ 2/3 }(1.5\times 10^{ 8 }km)\)
\(=14.3\times 10^{ 8 }km\) 

Given  
 \({ M }_{ e }=6\times { 10 }^{ 24 }kg,{ r }_{ e }=6400km\)  
\( { r }_{ 1 }=6400+200=6600km=6.6\times { 10 }^{ 6 }m\) 
\({ r }_{ 2 }=6400+199=6599km=6.599\times { 10 }^{ 6 }m\)
Change in energy =   \(GMm\left( \frac { 1 }{ { r }_{ 1 } } -\frac { 1 }{ { r }_{ 2 } } \right) \)
\(\\ =6.67\times `{ 10 }^{ -11 }\times 6\times { 10 }^{ 24 }\times 500\)
\( \left( \frac { 1 }{ 6.6\times { 10 }^{ 6 } } =\frac { 1 }{ 6.599\times { 10 }^{ 6 } } \right) \)
\(\\ =2\times { 10 }^{ 17 }(1.5152\times { 10 }^{ -7 }-1.5154\times { 10 }^{ 7 })J\)
\(=-4\times { 10 }^{ 6 }J\)
If this occurs during one orbit,then the energy lost = \(force\times distance\) .If we take the sistances as being the circumference of one orbit Then
 Retarding force
\(=\frac { 4\times { 10 }^{ 6 } }{ 2\pi \times 6.6\times { 10 }^{ 6 } } =\frac { 4\times { 10 }^{ 6 } }{ 2\times 6.6\times 3.14\times { 10 }^{ 6 } } =0.1N\) 

\(-6.84\times 10^{ 7 }Jkg^{ -1 }\)

Mass of the earth, \(M=6.0\times { 10 }^{ 24 }kg\)
Mass of the sarellite, \(m=200kg\)
Radius of the earth, \(R=6.54\times { 10 }^{ 6 }\quad m\)
Height of the satellite above the earth's surface,
\(h=400km=0.4\times { 10 }^{ 6 }m\)
Radius of the orbit of the satellite \(r=R+h\)
\(=6.4\times { 10 }^{ 6 }+0.4\times { 10 }^{ 6 }\)
\(=6.8\times { 10 }^{ 6 }m\)  
Total energy of the satellite,
  \(E=\frac { GMn }{ 2r } =\frac { 6.67\times { 10 }^{ -11 }\times 6.0\times { 10 }^{ 24 }\times 200 }{ 2\times 6.8\times { 10 }^{ 6 } } \)
\(=-5.9\times { 10 }^{ 9 }J\)
Negative total energy denoted that the satellite is round to the earth.there fore,to pull the satellite out of the earth's gravitational influence, energy required \(=-5.9\times { 10 }^{ 9 }J\)

Weight of the body at the earth's surface
w = mg = 250 N ... (i)
Acceleration due to gravity at depth d from the earth's surface
\({ g }^{ ' }=g\left( 1-\frac { d }{ R } \right) \)
Here, \(\quad d=\frac { R }{ 2 }\)
\( \\ \therefore g^{ ' }=g\left( 1-\frac { { R }/2 }{ R } \right) =g\left( 1-\frac { 1 }{ 2 } \right)\)
\( \Rightarrow g^{ ' }=\frac { g }{ 2 } \)
Weight of the body at depth d
\(\Rightarrow \quad { w }^{ ' }=mg=\frac { mg }{ 2 } \)
Using Eq.(i) we get
\({ w }^{ ' }=\frac { 250 }{ 2 } =125N\)
\(\therefore\) Weight of the body will be 125N.

Period of a revolution of a satellite is the time taken by the satellite to complete one revolution round the earth. It is denoted by T.
\(\therefore T=\frac { Circumference\ of\ circular\ orbit }{ Orbital\ velocity } \) 
or    \(T=\frac { 2\pi r }{ { v }_{ o } } \)
or     \(T=\frac { 2\pi (R+h) }{ { v }_{ o } } \quad \quad \quad \quad \quad [\therefore r=R+H]\)
or     \(T=2\pi (R+h)\sqrt { \frac { R+h }{ GM } } \left[ \because \quad { v }_{ o }=\sqrt { \frac { GM }{ R+h } } \right] \)
or     \(T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ GM } } \)     
Also,  \(T=2\pi \sqrt { \frac { (R+h)^{ 2 }(R+h) }{ GM } } \)
or       \(T=2\pi \sqrt { \frac { (R+h)^{ 3 } }{ gR^{ 2 } } } \)
 \(\because \quad \quad g{ R }^{ 2 }=GM\)  
\(\therefore T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ gR^{ 2 } } } \)   

Given mass of satelite m=400 kg
initial energy is given by \(E_i=\frac{-G M m}{4 R}\)
final energy is givne by \(E_f=\frac{-G M m}{8 R}\)
\( \triangle E=\frac{G M m}{8 R}=\frac{G M}{R^2} \frac{m R}{8} \)
\(=\frac{g m R}{8} \)
\(=\frac{9.81 \times 400 x 6.37 \times 10^6}{8} \)
\( \triangle E=3.13 \times 10^9 j\)
\(-3.13\times 10^{ 9 }J,\ -6.26\times 10^{ 9 }J\)

We are given that
Mass of the Mars. \(M=6.4\times { 10 }^{ 23 }kg\)
Radius of the mars \(R=3395km=3.395\times { 10 }^{ 6 }m\)
Velocity of the rocket, \(v=2km/s=2\times { 10 }^{ 3 }m/s\)
Let h be the maximum height attained by the rocket.Change in potential energy of rocket.
PE=final potential energy-initial potential energy
\(=-G\frac { Mn }{ (R+h) } -\left( -G\frac { Mm }{ R } \right)\)
\( \\ =-G\frac { Mn }{ (R+h) } +G\frac { Mm }{ R } \)
\(\\ =GMm\left( \frac { 1 }{ R } -\frac { 1 }{ R+h } \right) =GMm\frac { h }{ R(R+h) } \)
Here 20% of the kinetic energy of the rocket is lost due to Martian atmosphere
KE of the rocket which is converted into its potential energy
 \(=\frac { 80 }{ 100 } \times \frac { 1 }{ 2 } m{ v }^{ 2 }=0.4m{ v }^{ 2 }\)
Applying law of conservation of energy
 \(\Rightarrow GMm\frac { h }{ R(R+h) } =0.4{ mv }^{ 2 }\) 
\(\\ \Rightarrow GM\frac { h }{ { R }^{ 2 }+Rh } =0.4{ v }^{ 2 }\quad or\quad h=\frac { { R }^{ 2 } }{ \left( \frac { GM }{ 0.4{ v }^{ 2 }\quad } \right) -R } \)
\(\\ \Rightarrow h=\frac { 11.526\times { 10 }^{ 12 } }{ 26.68\times { 10 }^{ 6 }-3.395\times { 10 }^{ 6 } } m\)
\( \Rightarrow h=495\times { 10 }^{ 3 }m=495\ m\)

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
\( {{\left( \dfrac{{{T}_{neptune}}}{1year} \right)}^{2}}={{\left( \dfrac{4495.06\times {{10}^{6}}km}{149.6\times {{10}^{6}}km} \right)}^{3}} \)
\( \Rightarrow {{\left( \dfrac{{{T}_{neptune}}}{1} \right)}^{2}}={{(30.05)}^{3}} \)
\( \Rightarrow {{T}_{neptune}}={{(30.05)}^{\frac{3}{2}}}=164.72years \)

Here, mass of each star, M = 2 x \({ 10 }^{ 30 }\)kg
Radius of each star, r = \({ 10 }^{7 }\)m
Initial potential energy of the stars when they are \({ 10 }^{ 12 }\)m apart
\(=-\frac { GM\times M }{ 10^{ 12 } } =-\frac { GM^{ 2 } }{ 10^{ 12 } } \)[distance between two stars = \({ 10 }^{ 12 }m\)]
when the stars are just going to collide, the distance between their centres =twice the radius of each star = 2r = \(2\times { 10 }^{ 7 }m\)
Final potential energy of the stars when they about to collide 
= \(-G\frac { M\times M }{ 2\times 10^{ 7 } } =-\frac { GM^{ 2 } }{ 2\times { 10 }^{ 7 } } \)
Change in potential energy of stars
\(=-\frac { GM^{ 2 } }{ 10^{ 12 } } -\left( -\frac { GM^{ 2 } }{ 2\times { 10 }^{ 7 } } \right) =\frac { GM^{ 2 } }{ 2\times 10^{ 7 } } -\frac { GM^{ 2 } }{ 10^{ 12 } }\)
\( \\ \approx \frac { GM^{ 2 } }{ 2\times 10^{ 7 } } \left[ as\frac { { GM }^{ 2 } }{ 2\times 10^{ 7 } } <<\frac { GM^{ 2 } }{ 2\times 10^{ 7 } } \right] ...(i)\)
Let v be the speed of each star just before colliding.
Final KE of the stars = \(2\times \frac { 1 }{ 2 } Mv^{ 2 }=Mv^{ 2 }\)
initial KE of the stars = 0
(because when the stars are initially \({ 10 }^{ 12 }\)m apart, their speeds are negligible)
Change in KE of the stars = \({ Mv }^{ 2 }\)....(ii)
Using the law of conservation of energy, from eqs(i) and (ii), we get 
\(\frac { GM^{ 2 } }{ 2\times 10^{ 7 } } =Mv^{ 2 }\)
\(\\ or\ v=\sqrt { \frac { GM }{ 2\times 10^{ 7 } } } or\quad v=\sqrt { \frac { 6.67\times 10^{ -11 }\times (2\times 10^{ 30 }) }{ 2\times 10^{ 7 } } } \)
\(\\ or\ v=2.6\times 10^{ 6 }m/s\)