Gravitation Test

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Gravitation Test - 73

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Question 1
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A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is

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

B
$${ 8 }km{ s }^{ -1 }$$

C
$${ 3.2 }km{ s }^{ -1 }$$

D
$${ 1.414 }\times { 8 }km{ s }^{ -1 }$$

Solution

Given: $$R= 6400 km$$ $$g=9.8 m/s^2$$
Orbital velocity of space ship revolving in an orbit of radius r, $$v_o= \sqrt{\dfrac{gR^2}{r}}$$
For $$r= R$$ $$\implies v_o= \sqrt{gR}= \sqrt{9.8 \times 6400\times 1000} \approx 7.919 km/s$$
Escape velocity, $$v_e= \sqrt{2gR} = \sqrt{2(9.8) \times 6400\times 1000}= 11.2 km/s$$
Thus additional velocity imparted $$= 11.2- 7.919 \approx 3.29 km/s $$
Question 2
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Find approximately the third cosmic velocity $$v_3$$ in km/s i.e. the minimum velocity that has to be imparted to a body relative to the Earth's surface to drive it out of the Solar system. It is given that the rotation of the Earth about its own axis is to be neglected.

A
$$16.6$$

B
$$20$$

C
$$10$$

D
None of these

Solution

Assume first that the attraction of the earth can be neglected. Then the minimum velocity, that must be imparted to the body to escape from the Sun's pull, is, as in $$1.230$$, equal to,
$$(\sqrt 2-1)v_1$$
$$v_1^2=\gamma M_s/r$$,
$$r$$ is radius of the earth's orbit,
$$M_s$$ is mass of the Sun.
In the actual case near the earth, the pull of the Sun is small and does not change much over distances, which are several times the radius of the Earth. The velocity $$v_3$$ in question is that which overcomes the earth's pull with sufficient velocity to escape the Sun's pull. Thus
$$mv_3^2/2-\gamma M_E/R=m(\sqrt 2-1)^2v_1^2/2$$
$$R$$ is radius of the earth,
$$M_s$$ is mass of the earth.
$$v_1^2=\gamma M_E/R$$,
so, $$v_3=\sqrt{2v_2^2+(\sqrt 2-1)^2v_1^2}=16.6\ km/s$$
Question 3
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A particle would take time $$t_1$$ to move down a straight tube from the surface of earth (supposed to be homogeneous sphere) to its centre. If gravitational acceleration were to remain constant, time would be $$t_2$$. The ratio $$t/t'$$ will be

A
$$\displaystyle \frac{\pi}{2 \sqrt{2}}$$

B
$$\displaystyle \frac{\pi}{2}$$

C
$$\displaystyle \frac{2\pi}{3}$$

D
$$\displaystyle \frac{\pi}{\sqrt{3}}$$

Solution

Lets consider a distance of dr at a distance of r from center of earth.
now, $$dt_1 = \dfrac{dr}{-v}$$ [since the velocity is towards the center of earth]
Energy is conserved , Initial = final energy
So, $$ \dfrac{-GMm}{R} = \dfrac{mv^2}{2} + \dfrac{-GMm}{2R^3}(3R^2 - r^2)$$
solving this, we get v = $$ (\dfrac{GM(R^2 - r^2)}{R^3})^{0.5}$$

Integrating on both sides with limits from R to -R
$$ t_1 = \pi (\dfrac{R^3}{GM})^{0.5} = \pi \times (\dfrac{R}{g})^{0.5} $$
for $$ t_2$$ , use S = $$ ut + \dfrac{a(t)^{2}}{2}$$
So, $$ R = 0 + \dfrac{g(t)^2}{2} \rightarrow t = (\dfrac{2R}{g})^{0.5} $$
But, $$ t_2 = 2t = 2(\dfrac{2R}{g})^{0.5}$$
Taking ratio, $$ \dfrac{t_1}{t_2} = \dfrac{\pi \times (\dfrac{R}{g})^{0.5}}{2 (\dfrac{2R}{g})^{0.5}} = \dfrac{\pi}{8^{0.5}}= \dfrac{\pi}{2 \sqrt2}$$
Question 4
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Directions For Questions

At point inside the earth the statements of gravitational field need to be modified. If one could drill a hole to the center of the earth and measure the force of gravity on a body at various distances from the center, the force would be found to decrease as the center is approached, rather than increasing as 1/r$$^2$$. Qualitatively, it is easy to see why this should be so; as the body enters the interior of the earth (or other spherical body)some of the earth's mass is on the side of the body opposite from the center of the earth and pulls the body in the opposite direction. Exactly at the center of the earth, the gravitational force on the body is zero. The magnitude of the gravitational constant $$G$$ can be found experimentally by measuring the force of gravitational attraction between two bodies of known masses $$m_1$$ and $$m_2$$ at a known separation for bodies of moderate size the force is extremely small, but it can be measured with an instrument called a torsion balance, invented by the Rev. John Michell and first used for this purpose by Sir Henry Cavendish in 1798. The same type of instrument was also used by Coulomb for studying forces of electrical and magnetic attraction and repulsion.

...view full instructions

Which of the following is correct?

A
The value of g is constant throughout

B
$$\displaystyle g \propto \frac{1}{r^2}$$

C
$$g$$ is slightly less (by about 1%) when distance $$< 200 m$$

D
$$g$$ is slightly greater when distance $$< 200 m$$.

Solution

The gravitational field inside earth is given by

$$g=\dfrac { GM }{ { r }^{ 2 } } (\dfrac { { r }^{ 3 } }{ { R }^{ 3 } } )=\dfrac { GMr }{ { R }^{ 3 } } $$

Hence, at a small depth h,

$$g'=\dfrac { GM(R-h) }{ { R }^{ 3 } } $$

Thus,

$$\Delta g=\dfrac { GMh }{ { R }^{ 3 } } =g\dfrac { h }{ R } $$

$$\dfrac { \Delta g }{ g } =\dfrac { h }{ R } $$

$$\dfrac { \Delta g }{ g } =\dfrac { 200 }{ 6400000 } <1 percent$$

Answer is option C.
Question 5
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The largest and the shortest distance of the earth from sun are a and b, respectively. The distance of the earth from sun when it is at a point where perpendicular drawn from the sun on the major axis meets the orbit is

A
$$\displaystyle \frac{ab}{a + b}$$

B
$$\displaystyle \frac{ab}{2(a + b)}$$

C
$$\displaystyle \frac{2ab}{a+b}$$

D
$$\displaystyle \frac{a+b}{2ab}$$

Solution

$$\dfrac{y}{r}=1-ecos\theta$$ (here $$e$$ is the eccentricity)
At $$r=a$$, $$\theta=0$$
or
$$\dfrac{y}{a}=1-e$$
At $$r=b$$ $$\theta=180^o$$
or
$$\dfrac{y}{b}=1+e$$
Thus
$$\dfrac{y}{a}+\dfrac{y}{b}=2$$
or
$$y=\dfrac{2ab}{a+b}$$
Question 6
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The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $$0.5 m$$. If mean density of the Moon is two-third that of the earth and radius is one quarter that of the earth, the maximum vertical distance which he can jump on the Moon and the ratio of the time of duration of the jump on the Moon to hold on the earth are

A
$$3 m$$, $$6:1$$

B
$$6 m$$, $$3:1$$

C
$$3 m$$, $$1:6$$

D
$$6 m$$, $$1:6$$

Solution

The work that the astronaut can do is same on both earth and the moon.
Hence, $$mg_{e}h_{e}=mg_{m}h_{m}$$
$$\implies m(\dfrac{GM_{e}}{R_{e}^2})h_{e}=m(\dfrac{GM_{m}}{R_{m}^2})h_{m}$$
Using $$M=\rho \dfrac{4}{3}\pi R^3$$,
$$h_{m}=h_{e}(\dfrac{\rho_{e}R_{e}}{\rho_{m}R_{m}})=3m$$
Again since work done has be to same. initial kinetic energy would same at both the places, hence the initial velocity.
$$t_{jump}=\dfrac{2v_{0}}{g}$$
$$\implies \dfrac{t_{m}}{t_{e}}=\dfrac{\rho_{e}R_{e}}{\rho_{m}R_{m}}=6:1$$
Question 7
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If $$V_e$$ is the escape velocity of a body from a planet of mass $$M$$ and radius $$R$$. Then, the velocity of satellite revolving at height $$h$$ from the surface of planet will be

A
$$V = V_e \sqrt{\dfrac{R}{(R + h)}}$$

B
$$V = V_e \sqrt{\dfrac{2R}{(R + h)}}$$

C
$$V = V_e \sqrt{\dfrac{(R + h)}{R}}$$

D
$$V = V_e \sqrt{\dfrac{R}{ 2 (R + h)}}$$

Solution

The expression of escape velocity is $$\displaystyle V_e=\sqrt{\dfrac{2GM}{R}} ..................................(1)$$
Let $$V$$ be the velocity of satellite revolving at height $$h$$ from the surface of planet.
here, $$\displaystyle \dfrac{mV^2}{(R+h)}=\dfrac{GMm}{(R+h)^2}$$
or $$\displaystyle V=\sqrt{\dfrac{GM}{R+h}} .......................(2)$$
$$\displaystyle (2)/(1), V=V_e\sqrt{\dfrac{R}{2(R+h)}}$$
Question 8
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A tunnel is dug along a chord of the earth at a perpendicular distance $$\displaystyle \frac{R}{3}$$ from the earth's centre. Assume wall of the tunnel is frictionless. Find the force exerted by the wall on mass m at a distance x from the centre of the tunnel

A
$$\displaystyle \frac{mg \sqrt{\frac{R^2}{9} + x^2}}{R}$$

B
$$\displaystyle \frac{mg x}{\sqrt{R^2 / a + x^2}}$$

C
$$\displaystyle \frac{mg}{3}$$

D
$$\displaystyle \frac{mg x}{R}$$

Solution

Mass enclosed in a sphere of radius $$(r=R/3)$$ is
$$\dfrac{M}{\dfrac{4}{3}\pi R^3}.\dfrac{4}{3}\pi r^3=\dfrac{Mr^3}{R^3}$$
Thus force on a wall of mass m at a distance r from center=$$\dfrac{G(M.\dfrac{r^3}{R^3})m}{r^2}=\dfrac{GMmr}{R^3}$$
For $$r=R/3$$,
$$F=\dfrac{GMm}{3R^2}=\dfrac{mg}{3}$$
Question 9
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A person brings amass of $$1 kg$$ from infinity to a point A lnitially, the mass was at rest but it moves at a speed of $$3 m/s$$ as it reaches A. The work done by the person on the mass is $$- 5.5 J$$ The gravitational potential at $$A$$ is

A
$$-1J/kg$$

B
$$-4.5 J/kg$$

C
$$-5.5J /kg$$

D
$$-10J/kg$$

Solution

Using work energy theorem:
$$W_{person}+W_{gravity} = \Delta K.E.$$

$$ -5.5+m (V_A-V_{infinity}) = \dfrac{1}{2}mv^2- 0 = \dfrac{1}{2}\times 1 \times 3^2$$

$$-5.5+ V_A -0= 4.5$$

$$V_A= 10 J/Kg$$
Question 10
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Four equal masses (each of mass $$M$$) are placed at the corners of a square of side $$a$$. The escape velocity of a body from the centre $$O$$ of the square is

A
$$\displaystyle \sqrt[4] {\frac {2GM}{a}}$$

B
$$\displaystyle \sqrt {\frac {8\sqrt {2}GM}{a}}$$

C
$$\displaystyle \frac {4GM}{a}$$

D
$$\displaystyle \sqrt {\frac {4\sqrt {2}GM}{a}}$$

Solution

Potential energy of particle at the centre of square

$$\displaystyle =-4 \left ( \dfrac {GMm}{\dfrac {a}{\sqrt {2}}} \right )$$

$$\displaystyle \therefore -4\left ( \dfrac {GMm}{a/\sqrt {2}}\right )+\dfrac {1}{2}mv^2=0 \Rightarrow v^2 = \dfrac {8\sqrt {2}GM}{a}$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 73

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Gravitation Test - 73

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Question 1

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

$${ 8 }km{ s }^{ -1 }$$

$${ 3.2 }km{ s }^{ -1 }$$

$${ 1.414 }\times { 8 }km{ s }^{ -1 }$$

Solution

Question 2

$$16.6$$

$$20$$

$$10$$

None of these

Solution

Question 3

$$\displaystyle \frac{\pi}{2 \sqrt{2}}$$

$$\displaystyle \frac{\pi}{2}$$

$$\displaystyle \frac{2\pi}{3}$$

$$\displaystyle \frac{\pi}{\sqrt{3}}$$

Solution

Question 4

The value of g is constant throughout

$$\displaystyle g \propto \frac{1}{r^2}$$

$$g$$ is slightly less (by about 1%) when distance $$< 200 m$$

$$g$$ is slightly greater when distance $$< 200 m$$.

Solution

Question 5

$$\displaystyle \frac{ab}{a + b}$$

$$\displaystyle \frac{ab}{2(a + b)}$$

$$\displaystyle \frac{2ab}{a+b}$$

$$\displaystyle \frac{a+b}{2ab}$$

Solution

Question 6

$$3 m$$, $$6:1$$

$$6 m$$, $$3:1$$

$$3 m$$, $$1:6$$

$$6 m$$, $$1:6$$

Solution

Question 7

$$V = V_e \sqrt{\dfrac{R}{(R + h)}}$$

$$V = V_e \sqrt{\dfrac{2R}{(R + h)}}$$

$$V = V_e \sqrt{\dfrac{(R + h)}{R}}$$

$$V = V_e \sqrt{\dfrac{R}{ 2 (R + h)}}$$

Solution

Question 8

$$\displaystyle \frac{mg \sqrt{\frac{R^2}{9} + x^2}}{R}$$

$$\displaystyle \frac{mg x}{\sqrt{R^2 / a + x^2}}$$

$$\displaystyle \frac{mg}{3}$$

$$\displaystyle \frac{mg x}{R}$$

Solution

Question 9

$$-1J/kg$$

$$-4.5 J/kg$$

$$-5.5J /kg$$

$$-10J/kg$$

Solution

Question 10

$$\displaystyle \sqrt[4] {\frac {2GM}{a}}$$

$$\displaystyle \sqrt {\frac {8\sqrt {2}GM}{a}}$$

$$\displaystyle \frac {4GM}{a}$$

$$\displaystyle \sqrt {\frac {4\sqrt {2}GM}{a}}$$

Solution

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