Gravitational Fields - Class XII

In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.

height$=h$

A force of 10 N of gravittional force in CGS units is represented by 10 Dynes.

The force experienced by a $unit $ mass at a point in the gravitational field $=F _g=mg=1\times g=g $ $ Newton $

Gravitational intensity is nothing but the acceleration due to gravity(g)

Gravitational intensity is nothing but the acceleration due to gravity $=g=F _g/m$

The region of space surrounding a body in which another body experiences a force of gravitational attraction is called a gravitational field.

The speed of gravitational waves in the general theory of relativity is equal to the speed of light in vacuum.

Gravitational potential is defined as the Gravitational energy per unit mass of the object on the planet. Unit of energy is always joule in SI units. Hence that of potential will be Joule/kg.

The value of G in SI units is $6.67408\times {10^{-11}}$

The gravitational field is the region in which the gravitational force can be experienced or its presence can be felt. The intensity of gravitational field is the force acting on a unit mass of a body. 

The gravitational field is the region in which the gravitational force can be experienced or its presence can be felt. The intensity of gravitational force is the force acting on a unit mass of a body. The gravitational field is mathematically given by, gravitational force divided by the mass of the body. And the unit of the gravitational field is the ratio of the unit of force to that of mass, that is, $N/kg$.

Point A is the definition of the gravitational field. 

The gravitational force on an extended object can be treated as its center of gravity when acted on a single point. Also, it is uniform across its center of mass.

Strength of gravity $\cfrac { Gm }{ { r }^{ 2 } } $

Force if gravity on an object due to Earth always act towards the Earth as it is attractive in nature. Therefore, when an object us thrown up, the force of gravity acts towards the Earth. 

$V=-\int { Edx } =-\int { \cfrac { K }{ { x }^{ 3 } } dx=K\left( \cfrac { 1 }{ 2{ x }^{ 2 } }  \right)  } =\cfrac { K }{ { 2x }^{ 2 } } $

Gravitational intensity is a vector quantity related to the condition at any point under gravitational influence the measure of which is the gravitational force exerted upon a unit mass placed at the point in question. It is denoted by $g.$ 

Gravitational field is the force field that exists in the space around every mass or group of masses. Mathematically, it is F/m

• The gravitational field is the actually the force experimented by a unit mass due to a body of mass M.
  
• So when a unit mass is kept at a distance 'r' from mass 'M' then it experiences a force which is equal to $\dfrac{GM}{r^2}$[as m=1],which is nothing but the gravitational field of mass M.
• Its units are $\dfrac{N}{kg}$
• Hence the above statement is true.

Gravitational force $=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } ={ F } _{ 1 }$

According to newton's laws the time taken to fall is not related to mass

The gravitational field at the midpoint of BC is the vector sum of the fields from the three masses. Using the geometry of the equilateral triangle, the components cancel or add up to the calculated resultant.

The escape velocity is related to temperature by the formula v_rms = sqrt(3RT/M). Setting v_rms equal to the escape velocity (11.2 km/s) and solving for T gives approximately 10075 K.

G on earth=9.8

The force field is similar to gravitational field.
So, the path is parabolic or elliptical or hyperbolic.
The exact trajectory will depend on the masses and  energy of the particle.
Option(A) is equation of a straight line, it is ruled out.
Option(B) is equation for a circle and is also eqn of an ellipse with length of major axis equal to minor axis.
Option(C) is eqn of a parabola.
Hence, (D) is the best option.

$v=x^2+y^2+z^2$

The mass of moon is $1.2$% of that of earth. Hence $M _{moon}=0.012M _{earth}$

$T =$ Time period

Gravitational force between two masses = $\quad K\dfrac { m1m2 }{ { r }^{ 2 } } \quad \quad $

Where

K is universal gravitational constant $m _1$ and $m _2$ are the masses $r$ is the distance between the masses.

Now , the acceleration due to $ m1 $ being earth will be = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad \quad $

If the radius decreases and the mass remains same the acceleration will increase.

Gravitational field = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad $

where $K$ is universal gravitational constant $m _1$ is the mass $r$ is the distance. 

G= $ 6.67\times { 10 }^{ -11 }\dfrac { 1.9({ 10 }^{ 27 }) }{ { (7.1\times { 10 }^{ 7 }) }^{ 2 } } \quad \quad $ 

   =  $\quad 25N{ kg }^{ -1 }\quad $

If the moon were to experience only the gravitation of the earth only then the path would have been elliptical but in the present case moon experience the gravitational force of sun as well which distorts the path.

A. stronout feels weightless inside an artificial satellite because nothing is stopping their fall due to force of gravity.

We know that for an equilateral triangle the line joining the center of gravity to each vertex of the triangle are each at angle $120^0$ and we also know that the field by each mass will be along these lines joining the center and vertex.

The gravitational field lines are directed inwards towards a particle because at any point on the earth's field, a body will feel a force directed towards the center of the earth. The field lines becomes more spread out as the distance form the earth increases, which indicates the diminishing strength of the field

The strength of a gravitational field is given by the number of lines crowding at a point. This field strength is given by $GM/R^2$. Thus, it is proportional to the mass of the object

Larger the mass, more the field intensity and hence the number of lines of forces

Thus, the correct option is (b)

Number of lines of force is given by the flux $\phi=\int (g.dA)=(GM/R^2)4 \pi R^2 = 4 \pi GM$

Thus, number of lines of forces is proportional to M. 

So, $M _A= \rho (4 \pi R^3)/3$ AND $M _B=27 \rho (4 \pi [R/3]^3)/3=\rho (4 \pi R^3)/3$

The mass of both the planets are same AND hence the flux is also same for both the planets

The correct option is (a)

As there is no gravitational field in the shell, there are zero gravitational lines of force inside a spherically symmetric shell.