Physics Gravitation

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 = (1/2) x 250 N = 125 N

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 is given by

W' = m X g' = (4/9) mg = (4/9) x w = (4/9) x 63 N = 28 N

Distance of the Earth from the Sun, $r_e$ = 1.50 * 10⁸ km = 1.5 * 10¹¹ m

Time period of the Earth = $T_e$ , Time period of the Saturn $T_s$ = ${29.5T}_e$

Distance of Saturn from the Sun = $r_s$

From Kepler's 3rd law of planetary motion, we have T =( ${\frac{4{\pi }^2{r}^3}{GM})}^{1/2}$

For Saturn and Sun, we can write, $\frac{r_s^3}{r_e^3}$ = $\frac{T_s^2}{T_e^3}$

$r_s$ = $r_e({\frac{29.5T_e}{T_e})}^{2/3}$ = 1.5 * 10¹¹$*({\frac{29.5}{1})}^{2/3}$ = 1.4 X ${10}^{12}$ m

Orbital radius of the Earth around the Sun, r = 1.5 * 10⁸ km = 1.5 * 10¹¹ m

Time taken by the Earth to complete 1 revolution around the Sun, T = 1 year = 365.25 days = 365.25 $*$ 24 $*24*60*60s$

Universal gravitational constant, G = 6.67 $*{10}^{-11}$ N $m^2/{kg}^2$

Thus, mass of the Sun can be calculated as,

M = $\frac{4{\pi }^2{r}^3}{G{T}^2}$ = $\frac{4{*{\left(3.1416\right)}^2}^{}{*(1.5*{10}^{11})}^3}{6.67*{10}^{-11}{*(365.25*24*60*60)}^2}$ = 2.0 $*{10}^{30}$ kg

If the upper half of the spherical shell is cut out, then the net gravitational force acting on a particle at an arbitrary point P 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 an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Is linear speed, angular speed, kinetic and potential energy varies from point to point in the orbit.

(a) No

(b) No

(c) Yes

(d) No

(e) No

(f) Yes

(a) Escape velocity of a body from the Earth is given by the relation:

$v_{esc}$ = $\sqrt{2gR}$ …. (1)

Where g = acceleration due to gravity and R = radius of the Earth

So escape velocity is independent of the mass of the body.

(b) It does not depend on the location from where it is projected.

(c) Does not depend on the direction of projection

(d) Depends on the height of the location from where the body is launched.