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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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Posted by

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

$U_{h}=-\frac{GMm}{R+h}$

Using $GM=gR^2$

$U_{h}=-\frac{gR^{2}m}{R+h}$

$U_{h}=-\frac{mgR}{1+\frac{h}{R}}$

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

$R\rightarrow$ 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

$v_{e}=\sqrt{\frac{2GM}{R}}$

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

$\Rightarrow n=4$

Hence the correct option is (1).

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Posted by

avinash.dongre

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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Posted by

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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Posted by

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_______.

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

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Posted by

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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Posted by

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)