Gravitation

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 $\times {10}^{6}m$

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

Potential energy of the spaceship due

Yes, a body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_g$ = $\frac{GMm}{R^2}$ , where M = mass of the star = 2.5 $\times 2\times {10}^{30}$ = 5 $\times {10}^{30}$ kg

M = mass of the body, R = rad

Mass of the Earth, M = 6.0 $*{10}^{24}$ kg

Radius of the Earth, R = 6400 km = 6.4 $*{10}^6$ m

Height of geostationary satellite from the surface of the Earth, h = 36000 km = 3.6 $*{10}^7$ m

Gravitational potential energy due to Earth's gravity at height h:

= $\frac{-GM}{(R+h)}$

= $-\frac{6.67*{10}^{-11}*6.0*{10}^{24}}{3.6*{10}^7+6.4*{10}^6}$ = $-\frac{6.67*6.0*{10}^{13}}{4.24*{10}^7}$ = $-$ 9

Mass of each sphere, M = 100 kg

Separation between the spheres, r = 1 m

X is the midpoint between the spheres.

Gravitational force at point x will be zero. This is because gravitational force exerted by each spheres will act in opposite directions.

Gravitational potential at point x:

= $\frac{-GM}{\frac{r}{2}}$ $\frac{-GM}{\frac{r}{2}}$ =

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 co

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 $\times {10}^6$ m

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

Orbital

Escape velocity of the projectile on the Earth’s surface, $v_{esc}$ = 11.2 km/s = 11.2 ${\times 10}^3$ m/s

Projection velocity of the projectile = 3 $v_{esc}$

Mass of the projectile = m

Velocity of the projectile far away from the Earth = $v_f$

Total energy of the projectile on the Earth = $\frac{1}{2}$ m ${v_p}^2$&nb

Velocity of the rocket, v = 5 km/s = 5 $\times {10}^3$ 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 h

Weight of a body of mass m at Earth's surface, W = mg = 250 N

Body of mass m is located at depth, d = $\frac{1}{2}{R}_e$ , where $R_e$ is radius of the Earth

Acceleration due to gravity at depth g (d) is given by : g' = (1 - $\frac{d}{R_e}$ )g = (1/2)g

Weight of the body at depth d

W' = mg' = (1/2) mg = (1/2)W

Weight of the body, W = 63 N

Acceleration due to gravity at h from the Earth's surface is given by

g' = $\frac{g}{{\left(\frac{1+h}{R_e}\right)}^2}$ where g = acceleration due to gravity on the Earth's surface,  $R_e$ = Radius of the Earth. h = $\frac{R_e}{2}$

g' = $\frac{g}{{\left(\frac{1+R_e/2}{R_e}\right)}^2}$ = $\frac{g}{{\left(\frac{1+h}{2h}\right)}^2}$ = (4/9)g

Weight of the body of mass m at a height h i