Tag: representing a gravitational field

Questions Related to representing a gravitational field

Value of gravitational constant, $'G'$ is

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $6.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$

  2. $4.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$

  3. $6.674 08 \times 10^{11} m^3 kg^{-1} s^{-2}$

  4. $6.674 08 \times 10^{-11} m^3 kg^{-2} s^{2}$

Show answer
Correct Option: A
Explanation:

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

And the unit is $ Newton.metre^2/kg^2 $
Newton is $kg.m.s^{-2} $
Hence it becomes $ m^3.kg^{-1}.s^{-2} $

Gravitational field is 

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. directly proportional to square of the distance between two masses.

  2. inversely proportional to square of the distance between two masses.

  3. directly proportional to the distance between two masses.

  4. inversely proportional to the distance between two masses.

Show answer
Correct Option: B
Explanation:

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. 

Gravitational force is given by $Force = \dfrac{GMm} {R^2}$, 
where $G$ is gravitational constant, $M$ is the mass of the body which is creating the gravitational force, $m$ is the mass of the body which is undergoing gravitational force and $R$ is the distance between them. You can see that this force is inversely proportional to the square of the distance between the two masses. Gravitational field is given by gravitational force / mass of the 2nd body i.e. $m$. 
$Field = \dfrac{GM}{R^2}$. Hence it is inversely proportional to square of the distance between two masses.

The unit of gravitational field is 

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $\dfrac{N}{kg}$

  2. $\dfrac{N}{s}$

  3. $\dfrac{kg}{s^2}$

  4. $\dfrac{N}{kg m^2}$

Show answer
Correct Option: A
Explanation:

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, $N/kg$.

What is gravitational field?

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. A gravitational field is a region where any other body that has mass will experience a force of attraction.

  2. A gravitational field is a region where any other body that has mass will experience a force of repulsion.

  3. A gravitational field is a region where any other body that has mass will experience no force of attraction.

  4. A gravitational field is a region where any other body that has mass will experience no force of repulsion.

Show answer
Correct Option: A
Explanation:

Point A is the definition of the gravitational field. 

Gravitational force is always attractive in nature and hence there is no question of repulsion.

Which of the following option is/are correct?

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. If acting at a single point, the gravitational force on an extended object can be treated as its centre of gravity

  2. If the gravitational field is nonuniform across the object then it can be treated as its centre of mass.

  3. If acting at multiple point, the gravitational force on an extended object can be treated as its centre of gravity

  4. If the gravitational field is uniform across the object then it can be treated as its centre of mass.

Show answer
Correct Option: A,D
Explanation:

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.

Determine the gravitational force of two particle of mass $3kg$ and $7 kg$ separated by a distance $2m$.

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $35 \times 10^{-11} N$

  2. $25 \times 10^{-11} N$

  3. $5 \times 10^{-11} N$

  4. $3.5 \times 10^{-11} N$

Show answer
Correct Option: A
Explanation:
Gravitational force of two particle of mass $3kg$ and $7kg$ separated by a distance $2m$ is given by: 
$F=G\dfrac{Mm}{r^2} = 6.674× 10^{-11} \times \dfrac{ 3 \times 7}{2^2}=35 \times 10^{-11} N$

How far from the centre of the Moon is the Earth-Moon neutral point, where the Earth and the Moon's gravitational field strengths are equal in magnitude but opposite in direction?

$ M _E= 6.0 \times 10^{24} kg \ \ \  M _M = 7.4 \times 10^{22} kg$
The radius of Moon's orbit (assumed to be circular) is: $3.8\times 10^{8} m$.

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $3.8 \times 10^{2} m$

  2. $38 \times 10^{6} m$

  3. $38 \times 10^{4} m$

  4. $28 \times 10^{6} m$

Show answer
Correct Option: B
Explanation:

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

Let ${ g } _{ e }$ be the value of $Gm$ of earth $=4\times { 10 }^{ 14 }\cfrac { { m }^{ 3 } }{ { s }^{ 2 } } $
      ${ g } _{ m }$ be the value of $Gm$ of moon $=5\times { 10 }^{ 12 }\cfrac { { m }^{ 3 } }{ { s }^{ 2 } } $
Let $d\rightarrow$distance between earth and moon $=380\times { 10 }^{ 6 }m$
Let equilibirum point from earth be at a distanexe :- $x$
       $\cfrac { { g } _{ e } }{ { x }^{ 2 } } =\cfrac { { g } _{ m } }{ { \left( d-x \right)  }^{ 2 } } $
      ${ g } _{ e }\left( { x }^{ 2 }-2xd+{ d }^{ 2 } \right) ={ g } _{ m }{ x }^{ 2 }\ \left( { g } _{ e }-{ g } _{ m } \right) { x }^{ 2 }-{ 2dg } _{ e }x+{ d }^{ 2 }{ g } _{ e }=0$
Solving the quadratic formula
$x=\cfrac { { 2dg } _{ e }\pm \sqrt { 4{ d }^{ 2 }{ g } _{ e }^{ 2 }-{ 4d }^{ 2 }{ g } _{ e }\left( { g } _{ e }-{ g } _{ m } \right)  }  }{ 2\left( { g } _{ e }-{ g } _{ m } \right)  } =\cfrac { { dg } _{ e }\pm d\sqrt { { g } _{ e }{ g } _{ m } }  }{ { g } _{ e }-{ g } _{ m } } \approx 342\times { 10 }^{ 6 }m$
$\therefore$ Distance from moon :-$d-x=38\times { 10 }^{ 6 }m$

Gravitational field is directed

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. towards the earth

  2. away from earth

  3. has no direction

  4. in a specific direction making angle with earth

Show answer
Correct Option: A
Explanation:

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. 

Due to a mass distribution, the gravitational field is $\dfrac{k}{x^3} $ along x-axis where $k$ is a constant. If the gravitational potential is taken to be at infinity, then the gravitational potential at $x$ is 

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $\dfrac{k}{x}$

  2. $\dfrac{k}{2x^2}$

  3. $\dfrac{k}{x^4}$

  4. $\dfrac{k}{x^6}$

Show answer
Correct Option: B
Explanation:

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

The gravitational intensity is denoted by :

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $g$

  2. $G$

  3. $E$

  4. none of these

Show answer
Correct Option: A
Explanation:

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 $g.$ 

$g = GM/$$R^2$, where $M$ is the mass of earth and $R$ is the radius of earth.