Tags Physics Gravitation

Physics Gravitation

Get insights from 160 questions on Physics Gravitation, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Gravitation

Follow Ask Question

160

Questions

Discussions

Active Users

Followers

New answer posted

6 months ago

Physics Gravitation Class 11th

8.5 Let us assume that our galaxy consists of 2.5 * 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milk Way to be 10⁵ ly.

0 Follower 9 Views

Vishal Baghel

Contributor-Level 10

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

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

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

Diameter of the Milky Way, d = 10⁵ ly

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

1 ly = 9.46 $* 10^{15}$ m

r = 5 $* 10^{4} *$ 9.46 $* 10^{15}$ m = 4.73 $* 10^{20}$ m

As a star revolves around the galactic centre of the Milky Way, its time period is given by the relation: $τ$ = ( ${\frac{4 π^{2} r^{3}}{G M})}^{\frac{1}{2}}$ = ( ${\frac{4 * {3.14}^{2} {* 4.73}^{3} * 10^{60}}{6.67 * 10^{- 11} * 5 * 10^{41}})}^{\frac{1}{2}}$ = 1.12 $* 10^{16} s$ = 3.55 x $10^{8}$ years

0 0

New answer posted

6 months ago

Physics Gravitation Class 11th

8.4 IO, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 * 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Radius of the Orbit is given by the relation, $R_{j u}$ = 4.22 $* 10^{8}$ m

Mass is given by the relation: $M_{j}$ = $\frac{4 π^{2} R_{j u}^{3}}{G T_{j u}^{2}}$ ……(i)

Where $M_{j}$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth, $T_{e}$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth, $R_{e}$ = 1 AU = 1.496 x $10^{11}$ m

Mass of Sun is given as : $M_{s}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ …(ii)

$\frac{M_{s}}{M_{j}}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ $*$ $\frac{G T_{j u}^{2}}{4 π^{2} R_{j u}^{3}}$ = $\frac{R_{e}^{3}}{T_{e}^{2}}$ $*$ $\frac{T_{j u}^{2}}{R_{j u}^{3}}$ = ( ${\frac{1.496 x 10^{11}}{4.22 * 10^{8}})}^{3}$ $* ({\frac{1.769 x 24 x 60 x 60}{365.25 x 24 x 60 x 60})}^{2}$ = 1045.04

Hence it can be inferred that the mass of Jupiter is about one – thousandth that

...more

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

Radius of the Orbit is given by the relation, $R_{j u}$ = 4.22 $* 10^{8}$ m

Mass is given by the relation: $M_{j}$ = $\frac{4 π^{2} R_{j u}^{3}}{G T_{j u}^{2}}$ ……(i)

Where $M_{j}$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth, $T_{e}$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth, $R_{e}$ = 1 AU = 1.496 x $10^{11}$ m

Mass of Sun is given as : $M_{s}$ = $\frac{4 π^{2} R_{e}^{3}}{G T_{e}^{2}}$ …(ii)

Hence it can be inferred that the mass of Jupiter is about one – thousandth that of the Sun

less

0 0

New answer posted

6 months ago

Physics Gravitation Class 11th

8.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

0 Follower 15 Views

Vishal Baghel

Contributor-Level 10

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 Kepler's 3rd law of planetary motion, we can write:

( ${\frac{R_{p}}{R_{e}})}^{3}$ = ( ${\frac{T_{p}}{T_{e}})}^{2}$

$\frac{R_{p}}{R_{e}}$ = ( ${\frac{T_{p}}{T_{e}})}^{2 / 3}$ = ( ${\frac{1}{2})}^{2 / 3}$ = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth

0 0

New answer posted

6 months ago

Physics Gravitation Class 11th

8.2 Choose the correct alternative:

(a) Acceleration due to gravity increases/decreases with increasing altitude

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density)

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body

(d) The formula –G Mm(1/r₂ – 1/r₁) is more/less accurate than the formula mg(r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

(a) Decreases - Acceleration due to gravity at depth h is given by $g_{h}$ = (1 – $\frac{2 h}{R_{e}}$ )g, where $R_{e} = R a d i u s o f t h e e a r t h,$ 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 – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

...more

(b) Decreases – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

(c) Mass of the body – Acceleration due to gravity of body mass m is given by the relation g = $\frac{G M}{R^{2}}$ , where G = Universal gravitation constant, M = mass of the Earth and R = radius of the Earth. Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More – Gravitational potential energy of two points $r_{1}$ and $r_{2}$ distance away from the centre of the Earth is respectively given by:

V( $r_{1}$ ) = $-$ - $\frac{G m M}{r_{1}}$ and V( $r_{2}$ ) = $-$ - $\frac{G m M}{r_{2}}$

Difference in potential energy, V = V( $r_{2}$ ) $-$ V( $r_{1}$ ) = $- G m M (\frac{1}{r_{2}}$ $- \frac{1}{r_{1}}$ )

Hence this formula is more accurate than mg( $r_{2}$ - $r_{1}$ )

less

0 0

New answer posted

6 months ago

Physics Gravitation Class 11th

8.12 A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero? Mass of the sun = 2*10³⁰ kg, mass of the earth = 6*10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 * 10¹¹ m).

0 Follower 4 Views

Pallavi Pathak

Contributor-Level 10

8.12

Mass of the Sun, $M_{s}$ = $2 * 10^{30}$ kg, Mass of the Earth, $M_{e}$ = $6 * 10^{24}$ kg

Orbital radius, r = $1.5 * 10^{11}$ m

Let the mass of the rocket be, m

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton's law of gravitation, we can equate gravitational forces acting on the satellite P under the influence of the Sun and the Earth as:

$\frac{G_{m} M_{s}}{({r - x)}^{2}}$ = $G_{m} \frac{M_{e}}{x^{2}}$ or( ${\frac{r - x}{x})}^{2}$

= $\frac{M_{s}}{M_{e}}$ ( ${\frac{r - x}{x})}^{2}$ = $\frac{{2 * 10}^{30}}{{6 * 10}^{24}}$ ,

$\frac{r - x}{x}$ = 577.35 , r = 578.35x ,

x = $\frac{1.5 * 10^{11}}{578.35}$ = 2.59 $* 10^{8}$ m

0 0

New answer posted

6 months ago

Physics Gravitation NCERT Physics 11th

Q.8.1 Answer the following:

(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun's pull is greater than the moon's pull

(You can check thi

...more

0 Follower 1 View

Pallavi Pathak

Contributor-Level 10

Ans.8.1

(a) No. Unlike electrical forces, gravitational force is independent of the status of the objects.

(b) Yes, the size of the space station is large enough and the astronaut will detect the change in Earth's gravity.

(c) Tidal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.

0 0

New answer posted

7 months ago

Physics Gravitation Class 11

What is the concept of weightlessness in gravitation and when does it occur?

0 Follower 3 Views

Pallavi Pathak

Contributor-Level 10

When there is no normal force acting on a body, weightlessness is experienced, and the body free fall under gravity. The concept of gravity is still acting on the body, however, there is no weight sensation of any reaction force from any surface or ground. The weightlessness felt by the astronauts in an orbiting spacecraft is not because gravity is not present but because while they are moving forward, both, they and the spacecraft are in continuous free fall towards the Earth. The role of the normal force in defining weight can be understood from this phenomenon also how it is different from the actual gravitational force.

0 0

New answer posted

7 months ago

Physics Gravitation Class 11

What is the concept of weightlessness in gravitation and when does it occur?

0 Follower 5 Views