Gravitation Test

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Gravitation Test - 16

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Weekly Quiz Competition

Question 1
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The dependence of acceleration due to gravity $$g$$ on the distance $$r $$ from the centre of the earth, assumed to be a sphere of radius $$R$$ of uniform density is as shown in figures below. The correct figure is

A

B

C

D

Solution

The acceleration due to gravity at a depth d below surface of earth is
$$\displaystyle g' = \frac{GM}{R^2} \left( 1 - \frac{d}{R} \right) = g \left( 1 - \frac{d}{R} \right)$$
$$g' = 0$$ at $$d = R$$
i.e., acceleration due to gravity is zero at the centre of earth.
Thus, the variation in value $$g$$ with $$r$$ is for, $$r > R$$,
$$\displaystyle g' = \frac{g}{ \left( \displaystyle 1 + \frac{h}{R} \right)^2} = \frac{gR^2}{r^2} \Rightarrow g' \propto \frac{1}{r^2}$$
Here, $$R + h = r$$
For $$\displaystyle r < R, g' = g \left( 1 - \frac{d}{R} \right) = \frac{gr}{R}$$
Here, $$R - d = r$$ $$\Rightarrow g' \propto r$$
Therefore, the variation of $$g$$ with distance from centre of the earth will be as shown in the figure.
Question 2
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The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

A
$$d=2km$$

B
$$d=\dfrac{1}{2}km$$

C
$$d=1km$$

D
$$d=\dfrac{3}{2}km$$

Solution

Acceteration due to gravity at height h,
$$g_n=g_0 (1-\dfrac{2h}{R})$$ $$ h=1km$$
Acceleration due to gravity at depth d,
$$d_d = g_0\left(1-\dfrac{d}{R}\right)$$
$$g_h=g_d$$
$$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})$$
$$\rightarrow d=2h$$
$$= 2\times 1 km$$
$$d=2km$$
Question 3
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If $${v}_{e}$$ is escape velocity and $${v}_{o}$$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

A
$${v}_{o}={v}_{e}$$

B
$${v}_{e}=\sqrt {2{v}_{o}}$$

C
$${v}_{e}=\sqrt {2}{v}_{o}$$

D
$${v}_{o}=\sqrt {2}{v}_{e}$$

Solution

$${v}_{escape}=\sqrt {\cfrac{2GM}{R}}$$
$${v}_{orbital}=\sqrt {\cfrac{GM}{R}}$$
$${v}_{escape}=\sqrt {2}{v}_{o}$$
Question 4
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A particle of mass $$m$$ is thrown upwards from the surface of the earth, with a velocity $$u$$. The mass and the radius of the earth are, respectively, $$M$$ and $$R$$. $$G$$ is gravitational constant and $$g$$ is acceleration due to gravity on the surface of the earth. The minimum value of $$u$$ so that the particle does not return back to earth is

A
$$\sqrt{\displaystyle\frac{2GM}{R}}$$

B
$$\sqrt{\displaystyle\frac{2GM}{{R}^{2}}}$$

C
$$\sqrt{2g{R}^{2}}$$

D
$$\sqrt{\displaystyle\frac{GM}{{R}^{2}}}$$

Solution

$$GM = g{R}^{2}$$
$${V}_{e} = \sqrt{2gR} = \sqrt{2\displaystyle\frac{GM}{{R}^{2}}}R = \sqrt{\displaystyle\frac{2GM}{R}}$$.
Question 5
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Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by which of the following options ?

A

B

C

D

Solution

Acceleration due to gravity, $$g = \dfrac{GMr}{R^3}$$ for $$0\leq r\leq R$$
$$\implies$$ $$g \propto r$$ for $$0\leq r\leq R$$

Acceleration due to gravity, $$g = \dfrac{GM}{r^2}$$ for $$ r\geq R$$
$$\implies$$ $$g \propto \dfrac{1}{r^2} $$ for $$ r\geq R$$
Hence option C is correct.
Question 6
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A body of mass m is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

A
$$3 mg R$$

B
$$\dfrac {1}{3} mg R$$

C
$$2 mg R$$

D
$$\dfrac {2}{3} mg R$$

Solution
Question 7
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The change in the gravitational potential energy when a body of mass $$m$$ is raised to a height $$nR$$ above the surface of the earth is (here $$R$$ is the radius of the earth).

A
$$\left (\dfrac {n}{n + 1}\right )mgR$$

B
$$\left (\dfrac {n}{n - 1}\right )mgR$$

C
$$nmgR$$

D
$$\dfrac {mgR}{n}$$

Solution

Gravitational potential energy of mass $$m$$ at any point at a distance $$r$$ from the centre of earth is
$$U = -\dfrac {GMm}{r}$$
At the surface of earth $$r = R$$,
$$\therefore U_{s} = -\dfrac {GMm}{R} = -mgR \left (\because g = \dfrac {GM}{R^{2}}\right )$$
At the height $$h= nR$$ from the surface of earth
$$r = R + h = R + nR = R (1 + n)$$
$$\therefore U_{h} = -\dfrac {GMm}{R(1 + n)} = -\dfrac {mgR}{(1 + n)}$$
Change in gravitational potential energy is
$$\triangle U = U_{h} - U_{s} = -\dfrac {mgR}{(1 + n)} - (-mgR)$$
$$=-\dfrac {mgR}{1 + n} + mgR$$
$$= mgR\left (1 - \dfrac {1}{1 - n}\right ) = mgR\left (\dfrac {n}{1 + n}\right )$$.
Question 8
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The value of $$g$$ at a height of $$100km$$ from the surface of the Earth is nearly (Radius of the Earth $$=$$ 6400km) ($$g$$ on the surface of the Earth $$=$$ $$9.8m/s^{2}$$)

A
$$9.5 ms^{-2}$$

B
$$8.5 ms^{-2}$$

C
$$10.5 ms^{-2}$$

D
$$9.8 ms^{-2}$$

Solution

Acceleration due to gravity changes with the height from Earth's surface as:
$$g^{'} = g(1-\dfrac {2h}{R})$$
$$\Rightarrow g^{'} = 9.8 (1-\dfrac {2\times 100}{6400})$$
$$\Rightarrow g^{'} = 9.8(1-\dfrac {1}{32})=9.49 m/s^2$$
Question 9
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The escape velocity for a body projected vertically upwards from the surface of earth is $$11$$ km/s. If the body is projected at an angle of $$45^{0}$$ with the vertical, the escape velocity will be

A
$$11\sqrt{2}km/s$$

B
$$22$$ $$km / s$$

C
$$11$$ $$km / s$$

D
$$11/\sqrt{2}$$ $$km / s$$

Solution

Escape speed of a body from Earth's surface is given by: $$v_{min}= \sqrt {2gR}$$
This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.
So, irrespective of the angle of projection, escape speed of the body from Earth's surface remains constant i.e. $$\approx 11$$ km/s
Question 10
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If $$g$$ on the surface of the Earth is $$9.8\ ms^{-2}$$, its value at a height of $$6400 km$$ is: (Radius of the Earth $$= 6400km$$)

A
$$4.9\ ms^{-2}$$

B
$$9.8\ ms^{-2}$$

C
$$2.45\ ms^{-2}$$

D
$$19.6\ ms^{-2}$$

Solution

The expression for $$g$$ is given by:
$$g_R =\dfrac {GM}{R^2}$$
The value of $$g$$ changes according to
$$g_h=\dfrac {GM}{(R+h)^2}$$
Therefore, $$\displaystyle \dfrac{g_{h}}{g_R} = \dfrac{R^{2}}{(R+h)^{2}}$$
Here, $$h = R$$
So, $$\dfrac {g_h}{g_R}=(\dfrac {R}{R+h})^2$$
$$\Rightarrow \dfrac {g_h}{g_R}=(\dfrac {R}{R+R})^2$$
$$\Rightarrow g_h=g_R\times\dfrac {1}{4} = 9.8\times 0.25=2.45\ m/s^2$$