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Consider two solid sphers of radii R_1=1m,\: R_2=2m and masses M1 and M2, respectively. The gravitational field due to sphere  and  are shown. The value of \frac{M_1}{M_2} is:  
Option: 1 \frac{2}{3}

Option: 2 \frac{1}{6}

Option: 3 \frac{1}{2}

Option: 4 \frac{1}{3}

Considering Case 2 and case 1, one by one

and using I=\frac{GM}{R^2} \ \ \ \ (\text{Gravitational field on surface of sphere})

R-Radius of sphere

                                                            \\ 3=\frac{\mathrm{Gm}_{2}}{2^{2}} \\ \\ 2=\frac{\mathrm{Gm}_{1}}{1^{2}} \\ \\ \therefore \frac{3}{2}=\frac{1}{4} \frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \\ \\ \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\frac{1}{6}

So the correct option is 2.

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vishal kumar

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches a height of R(R=radius of r=earth), it ejects a rocket of mass \frac{m}{10} so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth):
Option: 1 \frac{3m}{8}\left ( u+\sqrt{\frac{5GM}{6R}} \right )^{2}

Option: 2  5m\left ( u^{2}-\frac{119}{200}\: \frac{GM}{R} \right )  

Option: 3  \frac{m}{20}\left ( u^{2}+\frac{113}{200}\: \frac{GM}{R} \right )

Option: 4 \frac{m}{20}\left ( u-\sqrt{\frac{2GM}{3R}} \right )^{2}



Gravitational Potential Energy (U) -

It is the amount of work done in bringing a body from  \infty  to that point against gravitational force.

  • It is Scalar quantity

  • SI Unit: Joule

  • Dimension : \left[ ML^{2}T^{-2}\right ]

  • Gravitational Potential energy at a point

             If the point mass M is producing the field


         Then gravitational force on test mass m at a distance r from M is given by F=\frac{GMm}{r^2}

        And the amount of work done in bringing a body from \infty to r

                   = W=\int_{\infty}^{r}\frac{GMm}{x^2}dx=-\frac{GMm}{r}

         And this is equal to gravitational potential energy

               So U=-\frac{GMm}{r}

U \rightarrow gravitational potential energy 

M \rightarrow Mass of source-body

m \rightarrow mass of test body

r \rightarrow distance between two

Note- U is always negative in the gravitational field because Force is attractive in nature.

  Means As the distance r increases U becomes less negative 

I.e U will increase as r increases

And for r=\infty, U=o which is maximum

  • Gravitational Potential energy of discrete distribution of masses

  U=-G\left [ \frac{m_{1}m_{2}}{r_{12}}+\frac{m_{2}m_{3}}{r_{23}}+\cdot \cdot \cdot \right ]

U \rightarrow Net Gravitational Potential Energy

r_{12},r_{23}\rightarrow The distance of masses from each other


  • Change of potential energy

if a body of mass m is moved from  r_{1 } to r_{2 }

Then Change of potential energy is given as

 \Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]


\Delta U \rightarrow change of energy

r_{1},r_{2}\rightarrow distances

If r_{1}>r_{2} then the change in potential energy of the body will be negative.

              I.e To decrease potential energy of a body we have to bring that body closer to the earth.

  • The relation between Potential and Potential energy

As U=\frac{-GMm}{r}=m\left [ \frac{-GM}{r} \right ]

But V=-\frac{GM}{r}

So U=mV

Where V\rightarrow Potential

U\rightarrow Potential energy

r\rightarrow distance


  • Gravitational Potential Energy at the center of the earth relative to infinity

           {U_{centre}=mV_{centre}}\\ {V_{centre}\rightarrow Potential\: at\: centre}

           U=m\left ( -\frac{3}{2}\frac{GM}{R} \right )

               m \rightarrow mass of body

            M \rightarrow Mass of earth


  • The gravitational potential energy at height 'h' from the earth's surface


Using GM=gR^2





U_{h}\rightarrow The potential energy at the height h

R\rightarrow Radius of earth




Before the rocket rejection


Apply energy conservation

\frac{1}{2}mu^2-\frac{GMm}{R}= \frac{1}{2}mV^2-\frac{GMm}{2R}\\\Rightarrow \frac{1}{2}mV^2=\frac{1}{2}mu^2--\frac{GMm}{2R}\\ \Rightarrow V=\sqrt{u^2-\frac{GM}{R}}

After the rocket rejection

Apply momentum conservation 

Along y-axis

mV=\frac{m}{10}V_3\Rightarrow V_3=10V

Along x-axis

\frac{9m}{10}V_1=\frac{m}{10}V_2\Rightarrow V_2=9V_1

And since V_1=V_{orbital}=\sqrt{\frac{GM}{2R}}

So Kinetic energy of the rocket

K=\frac{1}{2}*\frac{m}{10}*(V_2^2+V_3^2)=\frac{1}{2}*\frac{m}{10}*(\frac{81GM}{2R}+100(u^2-\frac{GM}{R}))\\ K=5m(u^2-\frac{119}{200}*\frac{GM}{R})

So option (3) is correct.

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Ritika Jonwal

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Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are \nu _{A} and \nu _{B}, respectively, then \frac{\nu _{A}}{\nu _{B}}=\frac{n}{4}. The value of n is :
Option: 1 4
Option: 2 1
Option: 3 2
Option: 4 3




\therefore \frac{v_{1}}{v_{2}}=\frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{2GM/2}{R/2}}}=1=\frac{n}{4}

\Rightarrow n=4


Hence the correct option is (1).

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The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 \times 10^{3} \mathrm{~km}. Find the mass of Mars.
\left\{\right.$ Given $\left.\frac{4 \pi^{2}}{\mathrm{G}}=6 \times 10^{11} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \mathrm{~kg}^{2}\right\}
Option: 1 5.96 \times 10^{19} \mathrm{~kg}
Option: 2 3.25 \times 10^{21} \mathrm{~kg}
Option: 3 7.02 \times 10^{25} \mathrm{~kg}
Option: 4 6.00 \times 10^{23} \mathrm{~kg}

\begin{gathered} F=\frac{G M m}{r^{2}}=\frac{m v^{2}}{r} \\ V=\sqrt{\frac{G M}{r}} \\ T=\frac{2 \pi r}{v}=\frac{2 \pi r^{3 / 2}}{\sqrt{G M}} \\ 450 \times 60=\frac{2 \pi}{\sqrt{G M}} r^{3 / 2} \end{gathered}

\begin{aligned} (450 \times 60)^{2} &=\frac{4 \pi^{2}}{G M}\left(9 \times 10^{6}\right)^{3} \\ \end{aligned}

\begin{aligned} M &=\frac{6 \times 10^{11} \times\left(9 \times 10^{6}\right)^{3}}{(450 \times 60)^{2}} \\ &=\frac{9^{3} \times 6 \times 10^{29}}{450\times 60\times 450\times 60} \\ M &=6 \times 10^{23} \mathrm{~kg} \end{aligned}


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vishal kumar

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Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :
Option: 1 \sqrt{\frac{G}{2 R^{3}}}
Option: 2 \frac{1}{2} \sqrt{\frac{G}{R^{3}}}
Option: 3 \frac{1}{2 R} \sqrt{\frac{1}{G}}
Option: 4 \sqrt{\frac{2 G}{R^{3}}}

\begin{aligned} &F_{2}=F_{1}=\frac{G m_{1} m_{2}}{(2 R)^{2}} \end{aligned}

\begin{aligned} & F_{2}=F_{1}=m_{1} R \omega^{2}=m_{2} R \omega^{2}\\ \end{aligned}

\begin{aligned} &\frac{G}{4 R^{2}}=R \omega^{2}\\ \end{aligned}

\begin{aligned} &\omega^{2}=\frac{G}{4 R^{3}} \quad \omega=\left(\frac{G}{4 R^{3}}\right)^{1 / 2}\\ \end{aligned}

\begin{aligned} &\rightarrow \omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}} \end{aligned}

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vishal kumar

Inside a uniform spherical shell:
(a) the gravitational field is Zero.
(b) the gravitational potential is Zero.
(c) the gravitational field is same everywhere.
(d) the gravitational potential is same everywhere.
(e) all of the above
Choose the most appropriate answer from the options given below:
Option: 1 \left ( a \right ),\left ( c \right )and\left ( d \right )only
Option: 2 \left ( a \right ),\left ( b \right )and\left ( c \right )only
Option: 3 \left ( b \right ),\left ( c \right )and\left ( d \right )only
Option: 4 \left ( e \right )only

For uniform spherical shell,
I_{inside}= O\left ( r< R \right )
I_{outside}= \frac{GM}{r^{2}}\left ( r> R \right )
I_{surface}= \frac{GM}{R^{2}}\left ( r\geq R \right )
Gravitational field is zero inside shell and gravitational potential is constant inside shell
The correct option is 1

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vishal kumar

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A body of mass (2M) splits into four masses [\mathrm{m}, \mathrm{M}-\mathrm{m}, \mathrm{m}, \mathrm{M}-\mathrm{m} ],which are rearranged to form a square as shown in the figure. The ratio of \frac{M}{m} for which, the gravitational potential energy of the system becomes maximum is x: 1. The value of x is_______.

Gravitation \: PE\: of \: system = U_{12}+U_{23}+U_{34}+U_{41}+U_{13}+U_{24}
U= \frac{-4Gm\left ( M-m \right )}{d}-\frac{Gm^{2}}{\sqrt{2}d}-\frac{G\left ( M-m \right )^{2}}{\sqrt{2}d}
\frac{dU}{dm}= 0\: for\: U_{max}
\frac{dU}{dm}= 0= \frac{-\Delta G}{d}\left [M-2m-\frac{2m}{\sqrt{2}} -\frac{2\left ( M-m \right )}{\sqrt{2}} \left ( -1 \right )\right ]
0= \left [M-2m-\frac{2m}{\sqrt{2}} +\frac{2\left ( M-m \right )}{\sqrt{2}} \right ]
0= \left [M-2m+\frac{2M-4m}{\sqrt{2}} \right ]
4m+2\sqrt{2}m= 2M+\sqrt{2}M
2\sqrt{2}\left (\sqrt{2}+1 \right )m= \sqrt{2}\left ( \sqrt{2}+1 \right )M
\frac{M}{m}= \frac{2}{1}= \frac{X}{1}
\therefore X= 2

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vishal kumar

A mass of 50 \mathrm{~kg} is placed at the centre of a uniform spherical shell of mass 100 \mathrm{~kg} ad radius 50 \mathrm{~m}. If the gravitational potential at a point, 25 \mathrm{~m} from the centre is \mathrm{V} \mathrm{kg} / \mathrm{m}. The value of V is :
Option: 1 +2G
Option: 2 -60 G
Option: 3 -4 G
Option: 4 -20 G

We know that, for spherical shell

V_{0}= \frac{-GM}{R}\\                             \left [0< r\leq R \right ]\\

But  V= \frac{-GM}{R}\: -\frac{GM}{r}\\         (Given)

       V= \frac{-G(100)}{50}-\frac{G(50)}{25}\\

       V= -4G

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vishal kumar

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The masses and radii of the earth and moon are \left(\mathrm{M}_{1}, \mathrm{R}_{1}\right)$ and $\left(\mathrm{M}_{2}, \mathrm{R}_{2}\right) respectively. Their centres are at a distance ' r ' apart. Find the minimum escape velocity for a particle of mass ' \mathrm{m} ' to be projected from the middle of these two masses :
Option: 1 \mathrm{V}=\sqrt{\frac{4 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 2 \mathrm{V}=\frac{1}{2} \sqrt{\frac{2 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 3 \mathrm{V}=\frac{1}{2} \sqrt{\frac{4 \mathrm{G}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}}
Option: 4 \mathrm{V}=\frac{\sqrt{2 \mathrm{G}}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)}{\mathrm{r}}

Point P is not midpoint between o_{1} & o_{2}
PE= \frac{-GM_{1}m}{\frac{r}{2}}+\left ( \frac{-GM_{2}m}{\frac{r}{2}} \right )
For minimum escape velocity,
TE= 0= PE+KE
0= \frac{-2GM_{1}m}{r}-\frac{2GM_{2}m}{r}+\frac{1}{2}mv^{2}_{e}
V_{e}^{2}= \frac{4GM_{1}}{r}+\frac{4GM_{2}}{r}
\rightarrow V_{e}=\sqrt{\frac{4G\left ( M_{1}+M_{2} \right )}{r}}
The correct option is (1)

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vishal kumar

If R_{E} be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ' r ' below and a height ' r ' above the earth surface is:
(Given: r<R_{E} )
Option: 1 1+\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}-\frac{r^{3}}{R_{E}^{3}}
Option: 2 1+\frac{r}{R_{E}}+\frac{r^{2}}{R_{E}^{2}}+\frac{r^{3}}{R_{E}^{3}}
Option: 3 1-\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}-\frac{r^{3}}{R_{E}^{3}}
Option: 4 1+\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}+\frac{r^{3}}{R_{E}^{3}}

g_{h}=\frac{GM}{\left ( R +h\right )^{2}}=\frac{GM}{\left ( R+r \right )^{2}}=\frac{GM}{R^{2}\left ( 1+\frac{r}{R} \right )^{2}}\\             ........(1)

g_{d}=g\left ( 1-\frac{d}{R} \right )=\frac{GM}{R^{2}}\left ( 1-\frac{r}{R} \right )\\

\frac{g_{d}}{g_{h}}=\left ( 1-\frac{r}{R} \right )\left ( 1+\frac{r}{R} \right )^{2}\\

      =\left ( 1-\frac{r}{R}\right )\left ( 1+\frac{2r}{R}+\frac{r^{2}}{R^{2}} \right )\\


\frac{g_{d}}{g_{h}}= 1+\frac{r}{R}-\frac{r^{2}}{R^{2}}-\frac{r^{3}}{R^{3}}

The correct option is (1)

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vishal kumar

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