Gravitation Test

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Gravitation Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The escape velocity of 10g body from the earth is $$11.2 km s^{-1}$$. Ignoring air resistance, the escape velocity of 10 kg of the iron ball from the earth will be

A
$$0.0112 km s^{-1}$$

B
$$0.112 km s^{-1}$$

C
$$11.2 km s^{-1}$$

D
$$0.56 km s^{-1}$$

Solution

The escape velocity of a body from the surface of the Earth is the minimum velocity required by a body to go out of the Earth.
The escape velocity of the Earth is given by:
$$v_{e}=\sqrt{\dfrac{GM}{R}}$$

Here, $$G$$ is the gravitational constant, $$M$$ is the mass of the Earth and $$R$$ is the radius of Earth.

Here, the esape velocity of a body from the Earth does not depend on the mass of the body. It remains constant for all the body. So, for iron ball of mass $$10 kg$$, the escape velocity will be $$11.2\ km/s$$.

Option $$(C)$$ is correct.
Question 2
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ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH
Which of the following statements is correct?

A
Acceleration due to gravity increases with increasing altitude.

B
Acceleration due to gravity increases with increasing depth.

C
Acceleration due to gravity increases with increasing latitude.

D
Acceleration due to gravity is independent of the mass of the earth.

Solution

Acceleration due to gravity at a altitude h above the earth's surface is $$g_{h}=\dfrac{gR_{E}^{2}}{\left(R_{E}+{h}\right)^{2}}$$ ...(i)
where g is the acceleration due to gravity on the earth's surface and R_{E} is the radius of the earth.
Eq. (i) shows that acceleration due to gravity decreases with increasing altitude.

Acceleration due to gravity at a depth d below the earth's surface is $$g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)$$ ....(ii)
Eq. (ii) shows that acceleration due to gravity decreases with increasing depth.

Acceleration due to gravity at latitude $$\lambda$$
$$g\lambda=g-R_{E}\omega^{2}\cos^{2}\lambda$$ .....(iii)
where $$\omega$$ is the angular speed of rotation of the earth.
Eq.(iii) shows that acceleration due to gravity increases with increasing latitude.

Acceleration due to gravity of body of mass m is placed n the earth's surface is $$g=\dfrac{GM_{E}}{R_{E}^{2}}$$ ....(iv)
Eq.(iv) shows that acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
Question 3
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ESCAPE SPEED
The escape speed velocity of a body from the earth depends on
(i) the mass of the body
(ii) the location from where it is projected.
(iii) the direction of projection
(iv) the height of the location from where the body is launched.

A
(i) and (ii)

B
(ii) and (iv)

C
(i) and (iii)

D
(iii) and (iv)

Solution

The escape velocity is independent of mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, the escape velocity depends slightly on these factors.
Question 4
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Which of the following statements is correct regarding the universal gravitational constant G?

A
G has same value in all systems of units.

B
The value of G is same everywhere in the universe.

C
The value of G was first experimentally determined by Johannes Kepler.

D
G is a vector quantity

Solution

G has different value in different system of units.
In SI system the value of G is $$6.67\times 10^{-11}N m^{2} kg^{-2}$$ whereas in CGS its value is $$6.67\times 10^{-8}dyne cm^{2} g^{-2}$$.
The value of G is same throughout the universe.
The value of G was first experimentally determined by English scientist Henry Cavendish.
G is a scalar quantity.
Question 5
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The escape velocity from the surface of the earth is (where $$R_{E}$$ is the radius of the earth)

A
$$\sqrt{2gR_{E}}$$

B
$$\sqrt{gR_{E}}$$

C
$$2\sqrt{gR_{E}}$$

D
$$\sqrt{3gR_{E}}$$

Solution

The escape velocity from the surface of the earth is
$$v_e=\sqrt{\dfrac{2GM}{R_E}}$$. . . . .(1)
Acceleration due to gravity,
$$g=\dfrac{GM}{R_E^2}$$
$$\dfrac{GM}{R_E}=gR_E$$. . . . . . . . . .(2)
Substitute equation (2) in equation (1), we get
$$v_e=\sqrt{2gR_E}$$
The correct option is A.
Question 6
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The areal velocity and the angular momentum of the planet are related by which of the following relations?
(where $${m}_{p}$$ is the mass of the planet)

A
$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{2m}_{p}}$$

B
$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{m}_{p}}$$

C
$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{2\overset{-}{L}}{{m}_{p}}$$

D
$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{\sqrt{2m}_{p}}$$

Solution

Lets consider,
$$m_p=$$ mass of the planet.
The angular momentum, $$L=m_pvr$$
$$vr=\dfrac{ L}{m_p}$$. . .. . .(1)
Time period, $$t=\dfrac{2\pi r}{v}$$
Total area swapped in time $$t$$ , $$A=\pi r^2$$
Areal velocity, $$\dfrac{A}{t}=\dfrac{\pi r^2}{2\pi r/ v}=\dfrac{vr}{2}$$
$$\dfrac{A}{t}=\dfrac{L}{2m_p}$$ (from equation 1)
In vector form, areal velocity is given by
$$\dfrac{\vec A}{t}=\dfrac{\vec L}{2m_p}$$
The correct option is A.
Question 7
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The dependence of acceleration due to gravity g on the distance r from the centers of the earth assumed to be a sphere of radius R of uniform density is as shown figure below.
The correct figure is

A
(i)

B
(ii)

C
(iii)

D
(iv)

Solution

The acceleration due to gravity at a depth d below the surface of earth is $${g}^{\prime}=g\left(1-\dfrac{d}{R}\right)=g\left(\dfrac{R-d}{R}\right)=g\dfrac{r}{R}$$ .....(i)
where R-d=r=distance of location from the centre of the earth. When r=0, $$g^{\prime}=0$$
From (i), $$g\propto{r}$$ till R=r, for which $$g^{\prime}=g$$
For $$r>R, g^{\prime}=\dfrac{gR^{2}}{\left(R+h\right)^{2}}=\dfrac{gR^{2}}{r^{2}}$$ or $$g^{\prime}\propto\dfrac{1}{r^{2}}$$
Here, $$ R + h = r $$
Therefore, the variation of g with distance r from centre of earth will be as shown in figure (4). Thus, option (d) is correct.
Question 8
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The acceleration due to gravity $$g$$ and density of the earth $$\rho$$ are related by which of the following relations.? (where G is the gravitational constant and $${R}_{E}$$ is the radius of the earth)

A
$$\rho=\dfrac{4\pi GR_{E}}{3g}$$

B
$$\rho=\dfrac{3g}{4\pi GR_{E}}$$

C
$$\rho=\dfrac{3G}{4\pi gR_{E}}$$

D
$$\rho=\dfrac{4\pi gR_{E}}{3G}$$

Solution

Acceleration due to gravity on earth is $$g=\dfrac{GM_{E}}{R_{E}^{2}}$$ ....(i)

As
$$\rho=\dfrac{M_{E}}{\dfrac{4}{3}\pi R_{E}^{3}} \Rightarrow {M}_{E}=\rho\dfrac{4}{3}\pi {R_{E}^{3}}$$

Substituting this value in Eq. (i), we get

$$g=\dfrac{G\left(\rho\dfrac{4}{3}\pi {R_{E}^{3}}\right)}{{R}_{E}^{2}}=\dfrac{4}{3}\pi\rho GR_{E}$$

or $$\rho=\dfrac{3g}{4\pi GR_{E}}$$
Question 9
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The escape velocity for a body projected vertically upwards from the surface of the earth is $$11.2$$ km $$s^{-1}$$. If the body is projected in a direction making an angle $$45^o$$ with the vertical, the escape velocity will be:

A
$$\dfrac{11.2}{\sqrt{2}}$$km $$s^{-1}$$

B
$$11.2\times \sqrt{2}$$km $$s^{-1}$$

C
$$11.2\times 2$$km $$s^{-1}$$

D
$$11.2$$km $$s^{-1}$$

Solution

$$V_e=\sqrt{2gr}$$
It doesn't depend upon direction of projection.
Question 10
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In order to shift a body of mass m from a circular orbit of radius $$3R$$ to a higher orbit of radius $$5R$$ around the earth, the work done is?

A
$$\dfrac{3GMm}{5R}$$

B
$$\dfrac{GMm}{2R}$$

C
$$\dfrac{2}{15}\dfrac{GMm}{R}$$

D
$$\dfrac{GMm}{5R}$$

Solution

Work done is shifting a body from an orbit of radius $$3R$$ to $$5R$$ is equal to change in potential energy of the body.
Potential energy $$=\dfrac { -{ GM }{ m } }{ r } $$
where $$M$$ is mass of the earth $$m$$ is mass of satellite $$r$$ is radius of orbit
$$W=$$ Final potential energy $$-$$ Initial potential energy
$$W=\dfrac { { -GM }{ m } }{ 5R } -\left( \dfrac { { -GM }{ m } }{ 3R } \right) $$
$$=\dfrac { { GM }{ m } }{ R } \left( \dfrac { 1 }{ 3 } -\dfrac { 1 }{ 5 } \right) =\dfrac { { 2GM }{ m } }{ 15R } $$

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Gravitation Test - 49

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Gravitation Test - 49

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SHARING IS CARING

Question 1

$$0.0112 km s^{-1}$$

$$0.112 km s^{-1}$$

$$11.2 km s^{-1}$$

$$0.56 km s^{-1}$$

Solution

Question 2

Acceleration due to gravity increases with increasing altitude.

Acceleration due to gravity increases with increasing depth.

Acceleration due to gravity increases with increasing latitude.

Acceleration due to gravity is independent of the mass of the earth.

Solution

Question 3

(i) and (ii)

(ii) and (iv)

(i) and (iii)

(iii) and (iv)

Solution

Question 4

G has same value in all systems of units.

The value of G is same everywhere in the universe.

The value of G was first experimentally determined by Johannes Kepler.

G is a vector quantity

Solution

Question 5

$$\sqrt{2gR_{E}}$$

$$\sqrt{gR_{E}}$$

$$2\sqrt{gR_{E}}$$

$$\sqrt{3gR_{E}}$$

Solution

Question 6

$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{2m}_{p}}$$

$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{m}_{p}}$$

$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{2\overset{-}{L}}{{m}_{p}}$$

$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{\sqrt{2m}_{p}}$$

Solution

Question 7

(i)

(ii)

(iii)

(iv)

Solution

Question 8

$$\rho=\dfrac{4\pi GR_{E}}{3g}$$

$$\rho=\dfrac{3g}{4\pi GR_{E}}$$

$$\rho=\dfrac{3G}{4\pi gR_{E}}$$

$$\rho=\dfrac{4\pi gR_{E}}{3G}$$

Solution

Question 9

$$\dfrac{11.2}{\sqrt{2}}$$km $$s^{-1}$$

$$11.2\times \sqrt{2}$$km $$s^{-1}$$

$$11.2\times 2$$km $$s^{-1}$$

$$11.2$$km $$s^{-1}$$

Solution

Question 10

$$\dfrac{3GMm}{5R}$$

$$\dfrac{GMm}{2R}$$

$$\dfrac{2}{15}\dfrac{GMm}{R}$$

$$\dfrac{GMm}{5R}$$

Solution

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