Tag: introduction to gravitation

Questions Related to introduction to gravitation

The gravitational force $F _{g}$ between two objects does not depend on

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. sum of the masses

  2. product of mass

  3. graviational constant

  4. distance between the masses

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Correct Option: A

The gravitational force between two points masses $m _{1}$ and $m _{2}$ at separation $r$ is given by $F=G\dfrac {m _{1}m _{2}}{r^{2}}$ The constant $G$

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. depends on system of unit only

  2. depends on media between masses only

  3. depends on both $a$ and $b$

  4. is independent of both $a$ and $b$

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Correct Option: D

At what height from the surface of the earth will the value of acceleration due to gravity be reduced by $36\%$ from the value at the surface?
(Radius of earth=$6400\ km$)

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $1500\ km$

  2. $1200\ km$

  3. $1000\ km$

  4. $1600\ km$

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Correct Option: D

Two electrons are a certain distance apart from one another. What is the order of magnitude of the ratio of the electric force between them to the gravitational force between them?

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $10^8 : 1$

  2. $10^{28} : 1$

  3. $10^{31} : 1$

  4. $10^{42} : 1$

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Correct Option: A

A cylinderical vessel is filled with equal amount of weight of mercury and water.The overall height of the two layer is 29.2 cm,specific gravity of mercury is 13.6.Then the pressure of the liquid at the bottom of the vessel is:

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $29.2 cm$ of water

  2. $\dfrac{29.2}{13.6} cm$ of mercury

  3. $4 cm$ of mercury

  4. $15.6 cm$ of mercury

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Correct Option: C
Explanation:
${ M } _{ Hg }={ M } _{ w }$
${ P } _{ Hg }\left( A\times { L } _{ Hg } \right) ={ P } _{ w }\left( A\times { L } _{ w } \right) $
$\therefore$   ${ L } _{ w }=13.6{ L } _{ Hg }$
${ L } _{ w }+{ L } _{ Hg }=29.2$
$\Rightarrow { L } _{ Hg }=2cm\quad \quad { L } _{ w }=27.2cm=2cm$ of $Hg$
Total pressure $=4cm$ of $Hg$.

Two balls, each of radius $R$, equal mass and density are placed in contact, than the force of gravitation between them is proportional to

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $F\propto \dfrac {1}{R^{2}} $

  2. $F\propto R $

  3. $F\propto R^{4} $

  4. $F\propto \dfrac {1}{R} $

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Correct Option: C
Explanation:
Given,

Two balls, each of radius $R$ and of equal mass and density, are placed in contact.

  • Step-1:

Find the Distance between the centre of two balls

Distance between the centre of two balls $=$ Sum of their Radii.

Distance between the centre of two balls $= R+R = 2R$.

  • Step-2:

Express the Mass of a ball as product of Density and Volume.

(This would be same for other ball ,given that two balls have equal mass)

Since, the shape of the ball is Sphere.

Volume of the ball $V=\dfrac 43 \pi R^3$

So,

Mass of the ball $m=\rho\times \dfrac 43 \pi R^3$

  • Step-3:

Find the force of gravitation between the two balls.

According to Newton's Law of Universal Gravitation

$F=\dfrac{GMm}{r^2}$

Where,

$F =$ Gravitational Force between two objects.

$G =$ Gravitational constant

$M =$ Mass of the first object

$m =$ Mass of the second object

$r =$ Distance between objects

Here,
.
$M=m=\rho \times \dfrac 43 \pi R^3$

Substituting Values

$\implies F=\dfrac{Gm^2}{(2R)^2}$

$\implies F=\dfrac{G(\rho \times \dfrac 43\pi R^3)^2}{4R^2}$

$\implies F=G\times \rho^2\times (\dfrac 43)^2 \times \dfrac 14\times \dfrac{R^6}{R^2}$

$\implies F=G\times \rho^2\times (\dfrac 43)^2 \times \dfrac 14 \times R^4$

$\implies F\propto R^4$

Therefore,

The force of gravitation between the two balls is proportional to $R^4$

A satellite of the earth is revolving in circular orbit with a uniform velocity V. If the gravitational force suddenly disappears, the satellite will

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. continue to move with the same velocity in the same orbit.

  2. move tangentially to the original orbit with velocity V.

  3. fall down with increasing velocity.

  4. come to a stop somewhere in its original orbit.

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Correct Option: B
Explanation:

The satellite is revolving around earth because the centripetal force is balanced by earth's gravitational pull.If the gravitational pull disappears, the satellite free of centripetal force. So, it will travel with its instantaneous velocity i.e. in the direction tangential to the circular path.

The gravitational force on a body of mass $5\ kg$ at the surface of the earth is $50\ N$. If earth is a perfect sphere, the gravitational force on a satellite of ass $200\ kg$ in a circuit orbit of radius same as diameter of the earth is.

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $200\ N$

  2. $400\ N$

  3. $500\ N$

  4. $800\ N$

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Correct Option: A

A body has a weight $90\ N$ on the earth's surface the mass of the moon is $1/9$ that of the earth's mass and its radius is $1/2$ that of the earth's radius. on the moon the weight of the body is :

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. $45\ N$

  2. $202.5\ N$`

  3. $90\ N$

  4. $40\ N$

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Correct Option: D

If the distance between two bodies is doubles, the force of gravitational attraction between them. 

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. Becomes four times

  2. Is doubled

  3. Is reduced to one-fourth

  4. Is reduced to half.

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Correct Option: C