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

Answers (1)

best_answer

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.

Posted by

Ritika Jonwal

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