Gravitation Test

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Gravitation Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If the radius of the earth were to shrink by $$2$$%, its mass remaining the same, the acceleration due to gravity on the earth's surface would

A
decrease by $$2$$%

B
increase by $$2$$%

C
increase by $$4$$%

D
decrease by $$4$$%

Solution

$$g=\cfrac { GM }{ { R }^{ 2 } } $$, if $$R$$ decreases, then $$g$$ increases
Taking log on both sides, we have
$$\log { g } =\log { G } +\log { M } -2\log { R } $$
Differentiating it we get
$$\cfrac { dg }{ g } =0+0-\cfrac { 2dR }{ R } $$
$$=-2\times \left( \cfrac { -2 }{ 100 } \right) =\cfrac { 4 }{ 100 } $$
$$=\cfrac { 4 }{ 100 } \times 100$$
$$=4$$%
Question 2
1 / -0

Kepler's laws of motion are applicable to _______

A
All objects

B
Planets

C
Objects on earth

D
Objects on moon

Solution

Kepler derived 3 laws to explain the motion planets but however he failed to establish a theory to explain it. Newton used these laws and concluded that the gravitational force of the sun exerted on the planets held them in their positions
Question 3
1 / -0

When an object is far from earth, 'g' is calculated by the formula:

A
$$G\frac{M\times m}{R^2}$$

B
$$G\frac{M}{R^2}$$

C
$$G\frac{M\times m}{d^2}$$

D
$$G\frac{M}{d^2}$$

Solution

When an object in on or near the surface of the earth, the expression for $$g$$ may be approximated as:
$$g=\dfrac{GM}{R^2}$$
But, for objects far away from the surface of the earth, $$g \propto \dfrac{1}{d^2}$$
Hence $$g=\dfrac{GM}{d^2}$$ would be the correct expression.
Question 4
1 / -0

Universal gravitational constant $$G$$ is proportional to distance $$r$$ between two masses as-

A
$$r^2$$

B
$$r$$

C
$$1/r$$

D
does not depend upon $$r$$

Solution

According to Newton's Law of gravitation, the gravitational force between two bodies of mass $$M$$ and $$m$$ separated by a distance $$r$$, is directly proportional to the product of the masses of the objects ( $$F \propto Mm$$) and inversely proportional to the square of distance between them $$F \propto \dfrac{1}{r^2}$$.

When we combine this information, we get
$$F \propto \dfrac{Mm}{r^2}$$
$$F= G \dfrac{Mm}{r^2}$$

where $$G$$ is the proportionality constant, called the Universal Gravitational constant.

Since $$G$$ is a constant it does not depend on $$r$$.
Question 5
1 / -0

Motion of planets in solar system is an example of conservation of :

A
Momentum

B
Velocity

C
Angular momentum

D
Energy

Solution

$$\textbf{Explanation:}$$
$$\bullet$$Kepler's second law or the law of equal areas says that the line joining the planet and sun sweeps the same area in equal intervals of time. This is to conserve angular momentum.
$$\bullet$$In the motion of planets around the sun the angular momentum is conserved. That is the total angular momentum of the system is constant. This is because there is no external torque acting on the system.
$$\textbf{Hence option C correct}$$
Question 6
1 / -0

Value of universal gravitational constant $$G$$ in CGS unit is-

A
$$6.67\times 10^{8}cm^3g^{1}s{}$$

B
$$6.67\times 10^{8}cm^3g^{-1}s^{-2}$$

C
$$6.67\times 10^{9}cm^3g^{1}s^{2}$$

D
$$6.67\times 10^{7}cm^3g^{-1}s^{2}$$

Solution

Newton's law of gravitation:
$$F=G \dfrac{Mm}{r^2}$$
$$G = \dfrac{F r^2}{Mm}$$
In CGS unit system:
r - distance - cm,
M,m - mass - grams,
F - force - dyne - $$g cm/s^2$$
Therefore units of G - $$\dfrac{(g \space cm/s^2)(cm^2)}{g^2} = cm^3g^{-1}s^{-2}$$
Question 7
1 / -0

Value of acceleration due to gravity of earth is maximum :

A
At centre of earth

B
At surface of earth

C
At a height of 50 km from earths surface

D
At a height of 12 km from earths surface

Solution

Acceleration due to gravity at a height $$h$$ above the surface $$g' = g\bigg( 1-\dfrac{2h}{R}\bigg)$$
This suggest that acceleration due to gravity decreases as we go up higher above the earth surface.
At the surface, we have $$h = 0$$
Thus acceleration due to gravity at the surface $$g' = g $$ (maximum).
Also acceleration due to gravity is zero at the centre of earth.
Thus acceleration due to gravity is maximum at the earth's surface.
Question 8
1 / -0

The ratio of the value of gravitational constant $$G$$ between Earth and Moon system and Earth and Sun system is:

A
Greater than $$1$$

B
Less than $$1$$

C
Equal to $$1$$

D
Can't be determined

Solution

According to Newton's law of gravitation, every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$$F=\propto \dfrac{Mm}{r^2}$$

$$F=G \dfrac{Mm}{r^2}$$
Gravitational constant $$G$$ is the proportionality constant which has fixed value $$6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$$.
Therefore, the ratio of the value of gravitational constant $$G$$ between Earth and Moon system and Earth and Sun system is equal to 1. (Because it has the same value)
Question 9
1 / -0

The gravitational potential is a

A
scalar

B
vector

C
scalar based on the mass of the particle

D
scalar or vector depending on the situation

Solution

The gravitational potential is a scalar

The correct option is (a)
Question 10
1 / -0

The value of gravitational acceleration 'g' at a height 'h' above the earth's surface is $$\dfrac{g}{4}$$ then (R = radius of earth)

A
$$h = R$$

B
$$h = \dfrac{R}{2}$$

C
$$h = \dfrac{R}{3}$$

D
$$h = \dfrac{R}{4}$$

Solution

Value of gravitational acceleration at a height $$h$$ above th earth's surface is given by $$g' = g\bigg(\dfrac{R}{R+h}\bigg)^2$$
$$\therefore$$ $$\dfrac{g}{4}= g\bigg(\dfrac{R}{R+h}\bigg)^2$$
Or $$\dfrac{R}{R+h} = \dfrac{1}{2}$$
Or $$h+R = 2R$$
$$\implies$$ $$h = R$$

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Gravitation Test - 22

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Gravitation Test - 22

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Question 1

decrease by $$2$$%

increase by $$2$$%

increase by $$4$$%

decrease by $$4$$%

Solution

Question 2

All objects

Planets

Objects on earth

Objects on moon

Solution

Question 3

$$G\frac{M\times m}{R^2}$$

$$G\frac{M}{R^2}$$

$$G\frac{M\times m}{d^2}$$

$$G\frac{M}{d^2}$$

Solution

Question 4

$$r^2$$

$$r$$

$$1/r$$

does not depend upon $$r$$

Solution

Question 5

Momentum

Velocity

Angular momentum

Energy

Solution

Question 6

$$6.67\times 10^{8}cm^3g^{1}s{}$$

$$6.67\times 10^{8}cm^3g^{-1}s^{-2}$$

$$6.67\times 10^{9}cm^3g^{1}s^{2}$$

$$6.67\times 10^{7}cm^3g^{-1}s^{2}$$

Solution

Question 7

At centre of earth

At surface of earth

At a height of 50 km from earths surface

At a height of 12 km from earths surface

Solution

Question 8

Greater than $$1$$

Less than $$1$$

Equal to $$1$$

Can't be determined

Solution

Question 9

scalar

vector

scalar based on the mass of the particle

scalar or vector depending on the situation

Solution

Question 10

$$h = R$$

$$h = \dfrac{R}{2}$$

$$h = \dfrac{R}{3}$$

$$h = \dfrac{R}{4}$$

Solution

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