Class 11th

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

$\frac{1}{2}{v}^2$ = $G{M}_e(\frac{1}{R_e}-\frac{1}{R_e+h})$ = $\frac{G{M}_{e}h}{R_e(R_e+h)}$ = $\frac{G{M}_e}{R_e^2}*\frac{R_{e}h}{(R_e+h)}$ = $\frac{{gR}_{e}h}{(R_e+h)}$ where g = $\frac{G{M}_e}{R_e^2}$ = 9.8 $\frac{m}{s^2}$

$v^2(R_e+h)$ = 2 ${gR}_{e}h$ $v^2{R}_e+v^{2}h$ = 2 ${gR}_{e}h$ h(2 ${gR}_e-v^2)$ = $v^2{R}_e$

h = $\frac{v^2{R}_e}{(2{gR}_e-v^2)}$ = 1.59 x ${10}^6$ m

Gravitational potential (V) is constant at all points in a spherical shell. Hence the gravitational gradient ( $\frac{dV}{dr})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric

If the upper half of a spherical shell is cut out then the net gravitational force acting on a particle located at the centre O will be in the downward direction

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c.

(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore swollen feet of an astronaut do not affect him/her in space

(b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears, nose and mouth constitute a person's face. These symptoms can affect astronaut in space

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space

(d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space