Physics Gravitation

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 do

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 ins

(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 appare

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

(a) Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (can be either positive or negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-Satellite system is a bound system, the total energy of the

Mass of our galaxy Milky Way, M = 2.5 $\times {10}^{11}$ solar mass

Solar mass = Mass of the Sun = 2.0 $\times {10}^{30}$ kg

Mass of our galaxy = 2.5 $\times {10}^{11}\times$ 2.0 $\times {10}^{30}$ = 5 $\times {10}^{41}$ kg

Diameter of the Milky Way, d = 10⁵ ly

Radius of the Milky Way = 5 $\times {10}^4$ ly

1 ly = 9.46 $\times {10}^{15}$ m

r = 5 $\times {10}^4\times$ 9.46&n

The rotation period of the satellite Io, $T_{ju}$ = 1.769 days = 1.769 x 24 x 60 x 60s

Radius of the Orbit is given by the relation, $R_{ju}$ = 4.22 $\times {10}^8$ m

Mass is given by the relation: $M_j$ = $\frac{4{\pi }^2{R}_{ju}^3}{G{T}_{ju}^2}$ ……(i)

Where $M_j$ = mass of Jupiter and G = Universal gravitational constant

O

Time taken by the Earth to complete one revolution around the Sun,  $T_e$ = 1 year

Orbital radius of the Earth in its orbit,  $R_e$ = 1 AU

Time taken by the planet to complete one revolution around the Sun,  $T_p$ = $\frac{1}{2}{T}_e$ = $\frac{1}{2}$ year

Orbital radius of the planet = $R_p$

From Ke

(a) Decreases - Acceleration due to gravity at depth h is given by $g_h$ = (1 – $\frac{2h}{R_e}$ )g, where $R_e=Radiusoftheearth,$ g = acceleration due to gravity on the surface of the Earth. From this equation, it is clear that acceleration due to gravity decreases with increase in height

(b) Decreases – Acceler