Gravitation Test

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Gravitation Test - 41

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Question 1
1 / -0

If $$g$$ is the acceleration due to gravity on the earth's surface, the gain of the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth will be :

A
$$2mgR$$

B
$$mgR$$

C
$$\dfrac{1}{2}mgR$$

D
$$\dfrac{1}{4}mgR$$

Solution

The potential energy of an object at the surface of the earth
$${ U }_{ 1 }=-\dfrac { GMm }{ R } $$ .......(i)
The potential energy of the object at a height $$h = R$$ from the surface of the earth
$${ U }_{ 2 }=-\dfrac { GMm }{ R+h } =-\dfrac { GMm }{ R+R } $$ .....(ii)
Hence, the gain in potential energy of the object
$$\Delta U={ U }_{ 2 }-{ U }_{ 1 }$$
$$\Delta U=-\dfrac { GMm }{ R+R } +\dfrac { GMm }{ R } $$
$$\Delta U=-\dfrac { GMm }{ 2R } +\dfrac { GMm }{ R } $$
$$\Delta U=\dfrac { 1 }{ 2 } \dfrac { GMm }{ R } $$
But we know that $$GM=g{ R }^{ 2 }$$
Hence, $$\Delta U=\dfrac { 1 }{ 2 } \dfrac { g{ R }^{ 2 }m }{ R } $$
or $$\Delta U=\dfrac { 1 }{ 2 } gRm$$
or $$\Delta U=\dfrac { 1 }{ 2 } mgR$$
Question 2
1 / -0

Calculate angular velocity of earth so that acceleration due to gravity at $$60^o$$ latitude becomes zero. (Radius of earth $$=6400$$km, gravitational acceleration at poles$$=10m{/s^2}, \cos 60^o=0.5$$)

A
$$7.8\times 10^{-2}rad/s$$

B
$$0.5\times 10^{-3}rad/s$$

C
$$1\times 10^{-3}rad/s$$

D
$$2.5\times 10^{-3}rad/s$$

Solution

We have, $$ { g }_{ 1 }=g-{ \omega }^{ 2 }R\cos ^{ 2 }{ \phi } $$
According to the question, $$\omega=\dfrac{\pi}{3}$$ and $$g_{1}=0$$.
So, we must have $$\omega =\sqrt { \dfrac { 4g }{ R } } =2.5\times { 10 }^{ -3 }rad/s$$
Question 3
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Assuming $${ g }_{ \left( moon \right) }=\left( \dfrac { 1 }{ 6 } \right) { g }_{ earth }$$ and $${ D }_{ \left( moon \right) }=\left( \dfrac { 1 }{ 4 } \right) { D }_{ earth }$$ where $$g$$ and $$D$$ are the acceleration due to gravity and diameter respectively, the escape velocity from the moon is:

A
$$\dfrac { 11.2 }{ 24 }\ { kms }^{ -1 }$$

B
$$11.2\times \sqrt { 24 } \ { kms }^{ -1 }$$

C
$$\dfrac { 11.2 }{ \sqrt { 24 } }\ { kms }^{ -1 }$$

D
$$11.2\times 24\ { kms }^{ -1 }$$

Solution

Escape velocity, $${ v }_{ e }=\sqrt { 2gR } $$

$$\Rightarrow \dfrac { { v }_{ m } }{ { v }_{ e } } =\sqrt { \dfrac { \dfrac { 1 }{ 6 } { g }_{ e }\times \dfrac { 1 }{ 4 } { R }_{ e } }{ { g }_{ e }\times { R }_{ e } } } =\dfrac { 1 }{ \sqrt { 24 } } $$

$$\Rightarrow { v }_{ m }=\dfrac { 11.2 }{ \sqrt { 24 } } \ { kms }^{ -1 }$$
Question 4
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Weight of a body of mass m decreases by $$1 \%$$ when it is raised to height $$h$$ above the earth's surface. If the body is taken to a depth $$h$$ in a mine, change in its weight is:

A
by $$0.5 \%$$ decrease

B
by $$2 \%$$ decrease

C
by $$0.5 \%$$ increase

D
by $$1 \%$$ increase

Solution

Hint:
The weight of a body is defined as the gravitational force exerted on the body.

Step 1: Calculate the density of the earth and write the formula of the gravitational force.

$$density=\dfrac{mass}{volume}$$
$$\rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}$$

Here, $$M$$ is the mass,
and $$R$$ is the radius of the earth.

The force between two masses can be expressed as,
$$F=\dfrac{GM_1M_2}{r^2}$$

Step 2: Find the value of $$h$$.
Let the mass of the earth be $$M$$ and the mass of the body is $$m$$.
The weight of the body on the earth is,
$$F=\dfrac{GMm}{R^2}$$

The weight of the body when it is at height $$h$$ is,
$$F_1=\dfrac{GMm}{(R+h)^2}$$

The weight is decreased by 1%, Hence,
$$\dfrac{F-F_1}{F} \times 100 =1$$
$$[1-\dfrac{\dfrac{GMm}{(R+h)^2}}{\dfrac{GMm}{R^2}}] \times 100 =1$$
$$\dfrac{(R+h)^2-R^2}{(R+h)^2}=\dfrac{1}{100}$$
$$\dfrac{R^2+h^2+2Rh-R^2}{(R+h)^2}=\dfrac{1}{100}$$
$$\because h<<R$$

$$\therefore {(R+h)}^2 \rightarrow R^2$$ and $$h^2$$ is very small, therefore it can be ignored.
$$\dfrac{2hR}{(R)^2}=\dfrac{1}{100}$$
$$h=\dfrac{R}{200}$$

Step 3: Finding the weight inside the earth at depth $$h$$
The weight of the body when it is at depth $$h$$ is,
$$F_2=\dfrac{GM_hm}{(R-h)^2}$$

Here $$M_h$$ is the effective mass of the earth at depth $$h$$.
$$mass=density\times volume$$
$$M_h=\rho \times \frac{4}{3}\pi (R-h)^3$$
$$M_h=\dfrac{M}{\dfrac{4}{3}\pi R^3} \times \frac{4}{3}\pi (R-h)^3$$
$$M_h=\dfrac{M}{R^3} \times(R-h)^3$$

Substituting this value in the expression of $$F_2$$
$$F_=\dfrac{GMm(R-h)}{(R)^3}$$

Step 4: Calculate the weight loss,
The weight loss can be calculated as,
$$\Delta W$$$$=\dfrac{F-F_1}{F} \times 100 $$
$$\Delta W$$$$=[1-\dfrac{\dfrac{GMm(R-h)}{(R)^3}}{\dfrac{GMm}{R^2}}] \times 100 $$
$$\Delta W$$$$=[1-{\dfrac{(R-h)}{(R)}}] \times 100 $$

Substituting the value of $$h$$,
$$\Delta W =0.5%$$%

Therefore, the weight decreased by $$0.5$$%.
$$\textbf{Hence option A correct}$$
Question 5
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How far above the earth's surface must an astronaut in space be if they are to feel a gravitational acceleration that is half what they would feel on the surface of the earth?

A
$${R}_{Earth}$$

B
$$2{R}_{Earth}$$

C
$$\dfrac{{R}_{Earth}}{2}$$

D
$$\sqrt{2}{R}_{Earth}$$

E
$$\sqrt{2}{R}_{Earth}-{R}_{Earth}$$

Solution

Let the radius of earth be $$R_{Earth}$$ and acceleration due to gravity at the surface be $$g$$.
Thus acceleration due to gravity at a height h above the surface $$g' = g\dfrac{R_{Earth}^2}{(R_{Earth } + h)^2} $$
$$\therefore$$ $$\dfrac{g}{2} = g\dfrac{R_{Earth}^2}{(R_{Earth } + h)^2} $$

OR $$R_{Earth} + h = \sqrt{2} R_{Earth}$$ $$\implies h = \sqrt{2}R_{Earth} - R_{Earth}$$
Question 6
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Calculate the value of $$g$$ on the surface of planet if the planet has $$1/500$$ the mass and $$1/15$$ the radius of the Earth.

A
$$0.3 m/s_2$$

B
$$1.6 m/s_2$$

C
$$2.4 m/s_2$$

D
$$4.5 m/s_2$$

E
$$7.1 m/s_2$$

Solution

The value of $$g$$ on the surface of planet$$=\dfrac{GM_{planet}}{R^2}$$
Value of $$g$$ on earth's surface$$=g=10m/s^2=\dfrac{GM_{earth}}{R_{earth}^2}$$
The value of $$g$$ on the planet$$=\dfrac{GM_{planet}}{R_{planet}^2}$$
$$=\dfrac{15^2}{500}\times g_{earth}=4.5m/s^2$$
Question 7
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An astronaut who weighs $$162$$ pounds on the surface of the earth is orbiting the earth at a height above the surface of the earth of two earth radii ($$h = 2R$$ where R is the radius of the earth.)
How much does this astronaut weigh while in orbit at this height (With how much force is the earth pulling on him while he is in orbit at this height?)

A
$$81\ pounds$$

B
$$40.5\ pounds$$

C
$$18\ pounds$$

D
$$54\ pounds$$

E
$$0\ pounds$$ (astronaut is weightless)

Solution

Given : $$h = 2R$$
Weight of astronaut on the earth surface $$W_s = 162$$ pounds
Thus weight of astronaut at height h $$W' = W_s \bigg( \dfrac{R}{R+h} \bigg)^2$$
$$\therefore$$ $$W' = 162 \bigg( \dfrac{R}{R+2R} \bigg)^2 =\dfrac{162}{9} = 18$$ pounds
Question 8
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A person normally weighs 800N at sea level, climbs to the top of a mountain. While on top of the mountain that person will weigh:

A
Zero

B
Approximately $$800N$$

C
Considerably less than $$800N$$ but more than zero

D
Considerably more than $$800N$$

E
Need more information

Solution

When we go up from sea level at a height $$h$$ gravitational acceleration is given by the relation $$g'=g\left(1-2h/R \right)$$ where R is the radius of earth , as in this case h is negligible compared to R therefore g will not change considerably and due to that weight $$mg$$ will decrease but negligibly by small amount. Therefore it will remain approximately $$800N$$.
Question 9
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Calculate the mass and weight of the object which is at height of twice the radius of the Earth from the surface of the Earth, if mass and weight of the object is $$m$$ and $$w$$ respectively.

A
its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{2}$$

B
its mass is $$m$$ and its weight is $$\dfrac{w}{8}$$

C
its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{4}$$

D
its mass is $$m$$ and its weight is $$\dfrac{w}{4}$$

E
its mass is $$m$$ and its weight is $$\dfrac{w}{9}$$

Solution

Weight of the object at the surface$$=w=\dfrac{GMm}{R^2}$$
Mass of the object is universally constant. Thus it remains $$m$$ everywhere.
The weight of the object at height $$h=2R$$,
$$F=\dfrac{GMm}{(h+R)^2}=\dfrac{GMm}{9R^2}=\dfrac{w}{9}$$
Question 10
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Fifteen joules of work is done on object A so that only its gravitational potential energy changes. Sixty joules of work is done on object B (same mass as object A) so that only its gravitational potential energy changes.
How many times does the height of object B change compared to the height change of object A, as result of the work done?

A
Object B changes height four times as much as object A changes height

B
Object B changes height sixteen times as much as object A changes height

C
Object B changes height two times as much as object A changes height

D
Object B changes height less than two times as much as object A changes height (but not the same amount)

E
Object B changes height the same amount as object A changes height

Solution

Since the work done on object is stored as potential energy of object,
$$W=mg\Delta h$$
$$\implies \dfrac{W_A}{W_B}=\dfrac{\Delta h_A}{\Delta h_B}=\dfrac{15}{60}=\dfrac{1}{4}$$
$$\implies \Delta h_B=4\Delta h_A$$

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Gravitation Test - 41

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Gravitation Test - 41

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Question 1

$$2mgR$$

$$mgR$$

$$\dfrac{1}{2}mgR$$

$$\dfrac{1}{4}mgR$$

Solution

Question 2

$$7.8\times 10^{-2}rad/s$$

$$0.5\times 10^{-3}rad/s$$

$$1\times 10^{-3}rad/s$$

$$2.5\times 10^{-3}rad/s$$

Solution

Question 3

$$\dfrac { 11.2 }{ 24 }\ { kms }^{ -1 }$$

$$11.2\times \sqrt { 24 } \ { kms }^{ -1 }$$

$$\dfrac { 11.2 }{ \sqrt { 24 } }\ { kms }^{ -1 }$$

$$11.2\times 24\ { kms }^{ -1 }$$

Solution

Question 4

by $$0.5 \%$$ decrease

by $$2 \%$$ decrease

by $$0.5 \%$$ increase

by $$1 \%$$ increase

Solution

Question 5

$${R}_{Earth}$$

$$2{R}_{Earth}$$

$$\dfrac{{R}_{Earth}}{2}$$

$$\sqrt{2}{R}_{Earth}$$

$$\sqrt{2}{R}_{Earth}-{R}_{Earth}$$

Solution

Question 6

$$0.3 m/s_2$$

$$1.6 m/s_2$$

$$2.4 m/s_2$$

$$4.5 m/s_2$$

$$7.1 m/s_2$$

Solution

Question 7

$$81\ pounds$$

$$40.5\ pounds$$

$$18\ pounds$$

$$54\ pounds$$

$$0\ pounds$$ (astronaut is weightless)

Solution

Question 8

Zero

Approximately $$800N$$

Considerably less than $$800N$$ but more than zero

Considerably more than $$800N$$

Need more information

Solution

Question 9

its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{2}$$

its mass is $$m$$ and its weight is $$\dfrac{w}{8}$$

its mass is $$\dfrac{m}{2}$$ and its weight is $$\dfrac{w}{4}$$

its mass is $$m$$ and its weight is $$\dfrac{w}{4}$$

its mass is $$m$$ and its weight is $$\dfrac{w}{9}$$

Solution

Question 10

Object B changes height four times as much as object A changes height

Object B changes height sixteen times as much as object A changes height

Object B changes height two times as much as object A changes height

Object B changes height less than two times as much as object A changes height (but not the same amount)

Object B changes height the same amount as object A changes height

Solution

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