Physics Gravitation

Initial kinetic energy of the rocket = $\frac{1}{2}{mv}^2$

Initial potential energy of the rocket = $\frac{-GMm}{R}$

Total initial energy = $\frac{1}{2}{mv}^2-\frac{GMm}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8 $*\frac{1}{2}{mv}^2$ $-\frac{GMm}{R}$

Maximum height reached by the rocket = h

At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

Total energy of the rocket at height h = $-\frac{GMm}{R+h}$

Applying the law of conservation of energy for the rocket, we can write:

0.4 $*{mv}^2$ $-\frac{GMm}{R}$ = $-\frac{GMm}{R+h}$

0.4 $v^2$ = GM( $\frac{1}{R}$ $-\frac{1}{R+h})$

0.4 $v^2$ = GM( $\frac{R+h-R}{R(R+h)}$ ) = $\frac{GMh}{R(R+h)}$

$\frac{R+h}{h}$ = $\frac{GM}{{0.4Rv}^2}$

$\frac{R}{h}$ + 1 = $\frac{GM}{{0.4Rv}^2}$ = $\frac{6.4*{10}^{23}*6.67*{10}^{-11}}{0.4*3395*{10}^3*{2000}^2}$

h = 495 km

Mass of the spaceship, $m_s$ = 1000 kg

Mass of the Sun, M = 2 * 10³⁰ kg

Mass of Mars, $m_m$ = 6.4*10²³ kg

Orbital radius of Mars, R = 2.28 *10⁸ km = 2.28 *10¹¹ m

Radius of Mars, r = 3395 km = 3.395 $*{10}^{6}m$

Universal Gravitational constant, G = 6.67*10^-11 N m2 kg–2

Potential energy of the spaceship due to the gravitational attraction of the Sun = $\frac{-GM{m}_s}{R}$

Potential energy of the spaceship due to the gravitational attraction of Mars= $\frac{-G{m}_s{m}_m}{r}$

Since the spaceship is stationed on Mars, its velocity and hence its kinetic energy will be zero

Total energy of the spaceship = $\frac{-GM{m}_s}{R}$ $-\frac{-G{m}_s{m}_m}{r}$ = $-G{m}_s(\frac{M}{R}$ + $\frac{m_m}{r}$ )

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system = - (Total energy of the spaceship)

= $G{m}_s(\frac{M}{R}$ + $\frac{m_m}{r}$ )

= 6.67 $*{10}^{-11}*{10}^3*(\frac{2*{10}^{30}}{2.28*{10}^{11}}$ + $\frac{6.4*{10}^{23}}{3.395*{10}^6})$

= 6.67 $*{10}^{-8}*(8.77*{10}^{18}$ + 1.88 $*{10}^{17}$ )

= 6.67 $*{10}^{-8}*8.958*{10}^{18}$

= 5.97 $*{10}^{11}$ J

Mass of each star, M = 2 * 10³⁰ kg, Radius of each star, R = 10⁴ km = ${10}^7$ m

Distance between stars, r = ${10}^9$ km = ${10}^{12}$ m

For negligible speed, v = 0

The total energy of two stars separated at a distance r

= $-\frac{GMM}{r}+\frac{1}{2}m{v}^2$ = $-\frac{GMM}{r}$ …….(i)

Now, consider the case when the stars are about to collide. Velocity of the stars = V, distance between the centres of the stars = 2R

Total kinetic energy of both stars = $\frac{1}{2}$ M $V^2$ + $\frac{1}{2}$ M $V^2$ = M $V^2$

Total potential energy of both stars = $-\frac{GMM}{2R}$

Total energy of two stars = M $V^2-\frac{GMM}{2R}$ …….(ii)

Using the law of conservation of energy, we can write

M $V^2-\frac{GMM}{2R}$ = $-\frac{GMM}{r}$

M $V^2=\frac{GMM}{2R}$ $-\frac{GMM}{r}$

$V^2$ = GM ( $\frac{1}{2R}$ $-$ $\frac{1}{r})$ = 6.67 $*{10}^{-11}*2*{10}^{30}$ ( $\frac{1}{2*{10}^7}-\frac{1}{{10}^{-12}}$ )

= 1.334 $*{10}^{20}*$ ( $\frac{1}{2*{10}^7}-\frac{1}{{10}^{-12}}$ )

V = 2.6 $*{10}^6$ m/s

Mass of the Earth, M = 6.0 * 10²⁴ kg

Mass of the satellite, m = 200 kg

Radius of the Earth, $R_e$ = 6.4 *10⁶ m

Universal gravitational constant, G = 6.67 * 10^–11 N m² kg^–2

Height of the satellite, h = 400 km = 0.4 $*{10}^6$ m

Total energy of the satellite at height h = $\frac{1}{2}{mv}^2$ + ( $\frac{-G{M}_{e}m}{R_e+h}$ )

Orbital velocity of the satellite, v = $\sqrt{\frac{G{M}_e}{R_e+h}}$

Total energy of the satellite at height h = $\frac{m}{2}$ ( $\frac{G{M}_e}{R_e+h})$ - ( $\frac{G{M}_{e}m}{R_e+h}$ ) = $-$ $\frac{1}{2}\left(\frac{G{M}_{e}m}{R_e+h}\right)\mathrm{}$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite $\mathrm{T}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}.$

Energy require to send the satellite out of its orbit = - (bound energy) = $\frac{1}{2}\left(\frac{G{M}_{e}m}{R_e+h}\right)$

= $\frac{1}{2}\left(\frac{6.67X{10}^{-11}*6*{10}^{24}*200}{{6.4*{10}^6}_{}+0.4*{10}^6}\right)$ = 5.88 $*{10}^9$ J

Velocity of the rocket, v = 5 km/s = 5 $*{10}^3$ m/s

Mass of the Earth, $M_e$ = 6.0 * 10²⁴ kg

Radius of the Earth, $R_e$ = 6.4 * 10⁶ m

Height reached by rocket mass, m = h

At the Earth's surface

Total energy of the rocket = Kinetic energy + Potential energy = $\frac{1}{2}$ m $v^2$ + ( $\frac{-G{mM}_e}{R_e}$ )

At highest point h, v = 0, and potential energy = $\frac{-G{mM}_e}{R_e+h}$

Total energy of the rocket at height h = $\frac{-G{mM}_e}{R_e+h}$

From the law of conservation of energy, we have,

Total energy of the rocket at Earth surface = Total energy at height h

$\frac{1}{2}$ m $v^2$ + ( $\frac{-G{mM}_e}{R_e}$ ) = $\frac{-G{mM}_e}{R_e+h}$ or $\frac{1}{2}{v}^2$ = $G{M}_e(\frac{1}{R_e}-\frac{1}{R_e+h})$