Gravitation — Page 9 | NEET Physics

Q81

Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is :

A 1 : 4

B 8 : 1

C 2 : 1

D 1 : 8

Correct Answer

Option B

Solution

Given, period of revolution of first satellite, T1 = 1h Period of revolution of second satellite, T2 = 8h $\therefore$

{{{T_1}} \over {{T_2}}} = {1 \over 8}

We know that,

\omega = {{2\pi } \over T}

\Rightarrow \omega \propto {1 \over T}

$\because$

{{{\omega _1}} \over {{\omega _2}}} = {{{T_2}} \over {{T_1}}} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {8 \over 1}

or

{\omega _1}:{\omega _2} = 8:1

Q82

A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is :

A

-

60 G

B +2 G

C

-

20 G

D

-

4 G

Correct Answer

Option D

Solution

{V_A} = \left[ { - {{G{M_1}} \over r} - {{G{M_2}} \over R}} \right]

= \left[ { - {{50} \over {25}}G - {{100} \over {50}}G} \right]

= - 4G

Q83

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.Reason (R): For a central force field the angular momentum is a constant.In the light of the above statements, choose the most appropriate answer from the options given below:

A (A) is not correct but (R) is correct

B Both (A) and (R) are correct but (R) is not the correct explanation of (A)

C Both (A) and (R) are correct and (R) is the correct explanation of (A)

D (A) is correct but (R) is not correct

Correct Answer

Option C

Solution

The assertion (A) states that the “radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time” which is just another way of stating Kepler's second law.

This implies that the areal velocity is constant.

The reason (R) states that "For a central force field the angular momentum is a constant."

In a central force system (like that of a planet orbiting the Sun under gravity), the force always acts along the line joining the two bodies.

This means there is no torque acting on the planet, and thus, its angular momentum is conserved.

The connection between angular momentum and areal velocity is given by:

\text{Areal velocity} = \frac{1}{2}r^2\dot{\theta} = \frac{L}{2m}

where

L

is the angular momentum and

m

is the mass of the planet. Since the angular momentum

L

is constant, the areal velocity must also remain constant.

Therefore, both the assertion and the reason are correct, and the reason correctly explains why the areal velocity (as stated in the assertion) is constant.

So, the most appropriate answer is: Option C: Both (A) and (R) are correct and (R) is the correct explanation of (A).

Q84

Two spherical bodies of mass

M

and

5M

& radii

R

&

2R

respectively are released in free space with initial separation between their centers equal to

12R

. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

A

2.5

R

B

4.5

R

C

7.5

R

D

1.5

R

Correct Answer

Option C

Solution

Let

t

be the time taken for the two masses to collide and

{x_{5M,}}\,{x_M}

be the distance travelled by the mass

5M

and

M

respectively.

The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,

F = {{GM \times 5M} \over {{{\left( {12R - x} \right)}^2}}}

Acceleration of mass M,

{a_M} = {{G \times 5M} \over {{{\left( {12R - x} \right)}^2}}}

Acceleration of mass 5M,

{a_{5M}} = {{GM} \over {{{\left( {12R - x} \right)}^2}}}

For mass

5M

u = 0,\,\,S = {x_{5M}},\,\,t = t,\,\,a{ = a_{5M}}

S = ut + {1 \over 2}a{t^2}

$\therefore$

{x_{5M}} = {1 \over 2}{a_{5M}}{t^2}\,\,\,\,\,\,\,\,\,\,...\left( {i} \right)

For mass

M

u = 0,\,\,s = {x_M},\,\,t = t,\,\,a = {a_M}

$\therefore$

s = ut + {1 \over 2}a{t^2} \Rightarrow

{x_M} = {1 \over 2}{a_M}{t^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Dividing

(ii)

by

(iii)

{{{x_{5M}}} \over {{x_M}}} = {{{1 \over 2}{a_5}_M{t^2}} \over {{1 \over 2}{a_M}{t^2}}}

= {{{a_{5M}}} \over {{a_M}}} = {1 \over 5}

$\therefore$

5{x_{5M}} = {x_M}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left....( {iii} \right)

From the figure,

{x_{5M}} + {x_M} = 12R - 2R - R = 9R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( iv \right)

From

(iii)

and

(iv)

{{{x_M}} \over 5} + {x_M} = 9R

$\therefore$

6{x_M} = 45R

$\therefore$

{x_M} = {{45} \over 6}R = 7.5R

So two sphere collide when the sphere of mass M covered the distance of 7.5R.

Q85

A satellite of mass

m

revolves around the earth of radius

R

at a height

x

from its surface. If

g

is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

A

{{g{R^2}} \over {R + x}}

B

{{gR} \over {R - x}}

C

{gx}

D

{\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}

Correct Answer

Option D

Solution

Gravitational force applied on the satellite, =

{{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,

For satellite, gravitational force = centripetal force $\therefore$

{{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,

where

v

is the orbital speed of satellite. also

\,\,g = {{GM} \over {{R^2}}}

\Rightarrow GM = g{R^2}

$\therefore$

{v^2} = {{g{R^2}} \over {R + x}}

\Rightarrow v = {\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}

Q86

If

\mathrm{R}

is the radius of the earth and the acceleration due to gravity on the surface of earth is

g=\pi^2 \mathrm{~m} / \mathrm{s}^2

, then the length of the second's pendulum at a height

\mathrm{h}=2 R

from the surface of earth will be, :

A

\dfrac{1}{9} \mathrm{~m}

B

\dfrac{8}{9} \mathrm{~m}

C

\dfrac{2}{9} \mathrm{~m}

D

\dfrac{4}{9} \mathrm{~m}

Correct Answer

Option A

Solution

To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$ , is related to the gravitational acceleration, $g$ , and the period, $T$ , by the formula:

T = 2\pi\sqrt{\frac{L}{g}}

Since we are talking about a second's pendulum, the period, $T$ , is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T = 2 \text{ seconds}$ .

Now let's find the gravitational acceleration at height $h = 2R$ where $R$ is the radius of the earth.

The general formula for gravitational acceleration at a height $h$ above the surface is:

g_h = \frac{g}{{\left(1 + \frac{h}{R}\right)}^2}

Plugging $h = 2R$ into the formula, we get:

g_h = \frac{g}{{(1 + \frac{2R}{R})}^2}

g_h = \frac{g}{{(1 + 2)}^2}

g_h = \frac{g}{3^2} = \frac{g}{9}

So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth.

Given that $g = \pi^2 \text{ m/s}^2$ , we get:

g_h = \frac{\pi^2}{9} \text{ m/s}^2

Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$ :

2 = 2\pi\sqrt{\frac{L_h}{g_h}}

1 = \pi\sqrt{\frac{L_h}{g_h}}

\frac{1}{\pi} = \sqrt{\frac{L_h}{g_h}}

Squaring both sides, we get:

\frac{1}{\pi^2} = \frac{L_h}{g_h}

Multiplying both sides by $g_h$ gives us the length $L_h$ :

L_h = \frac{g_h}{\pi^2}

Substituting $g_h$ into the equation yields:

L_h = \frac{\pi^2}{9\pi^2}

L_h = \frac{1}{9} \text{ m}

Therefore, the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth is $\dfrac{1}{9}$ meters.

The correct answer is Option A.

Q87

Given below are two statements: Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity (g) Statement II : The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum. In the light of the above statements, choose the correct answer from the options given below

A Statement I is false but statement II is true

B Statement I is true but statement II is false

C Both Statement I and Statement II are true

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

Let's analyze both statements: Statement I: Rotation of the earth shows an effect on the value of acceleration due to gravity (g).

This is true.

The value of $g$ is affected by the rotation of the Earth.

The centripetal force due to the Earth's rotation causes a reduction in the perceived gravitational acceleration for objects on the surface.

This effect is zero at the poles and increases towards the equator.

The formula for the effective acceleration due to gravity at a latitude $\theta$ is:

g_{\text{effective}} = g - R\omega^2\cos^2(\theta)

where: $g$ is the theoretical acceleration due to gravity without Earth's rotation, $R$ is the radius of the Earth, $\omega$ is the angular velocity of the Earth's rotation, and $\theta$ is the latitude.

Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.

This is false.

For clarification, the effect of the Earth's rotation on the value of $g$ is maximum at the equator and minimum at the poles.

At the equator, the centrifugal force is highest because of the maximum velocity due to Earth's rotation, which decreases the effect of gravity more than at any other latitude.

At the poles, the centrifugal force is zero since there is no rotational velocity contributing to a centripetal effect; therefore, the effect of rotation on $g$ is minimum (zero).

With these considerations, the correct option is: Option B: Statement I is true but Statement II is false.

Q88

A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass

{m \over 2}

collides with A with a velocity which is half

\left( {{{\overrightarrow v } \over 2}} \right)

the instantaneous velocity

{\overrightarrow v }

of A. The collision is completely inelastic. Then, the combined body :

A starts moving in an elliptical orbit around the planet.

B Falls vertically downwards towards the planet

C Escapes from the Planet's Gravitational field.

D continues to move in a circular orbit

Correct Answer

Option A

Solution

Orbital speed for of A is v =

\sqrt {{{GM} \over R}}

After collision, let the combined mass moves with speed v1 $\therefore$ mv +

{m \over 2}{v \over 2}

=

\left( {{{3m} \over 2}} \right){v_1}

$\Rightarrow$ v1 =

{{5v} \over 6}

Since after collision, the speed is not equal to orbital speed at that point.

So motion cannot be circular.

Since velocity will remain tangential, so it cannot fall vertically towards the planet.

Their speed after collision is less than escape speed

\sqrt 2 v

, so they cannot escape gravitational field. So their motion will be elliptical around the planet.

Q89

Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are vA and vB, respectively, then

{{{v_A}} \over {{v_B}}} = {n \over 4}

. The value of n is :

A 1

B 2

C 4

D 3

Correct Answer

Option C

Solution

Escape velocity, Ve =

\sqrt {{{2GM} \over R}}

VA =

\sqrt {{{2GM} \over R}}

VB =

\sqrt {{{2G{M \over 2}} \over {{R \over 2}}}}

=

\sqrt {{{2GM} \over R}}

{{{V_A}} \over {{V_B}}}

= 1 Given

{{{V_A}} \over {{V_B}}} = {n \over 4}

$\therefore$

{n \over 4}

= 1 $\Rightarrow$ n = 4

Q90

The acceleration due to gravity at height

h

above the earth if $$h

A

g^{\prime}=g\left(1-\dfrac{2 h}{R}\right)

B

g^{\prime}=g\left(1-\dfrac{2 h^{2}}{R^{2}}\right)

C

g^{\prime}=g\left(1-\dfrac{h^{2}}{2 R^{2}}\right)

D

g^{\prime}=g\left(1-\dfrac{h}{2 R}\right)

Correct Answer

Option A

Solution

The acceleration due to gravity ( $g$ ) at the surface of Earth is approximately $9.81 \, \text{m/s}^2$ .

This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth.

This force decreases with the square of the distance from the center of the Earth, due to the inverse square law.

However, if we are at a height ( $h$ ) much less than the radius of the Earth ( $R$ ), we can use a linear approximation to calculate the new acceleration due to gravity ( $g'$ ) at this height.

This is because for small heights compared to the radius of Earth, the decrease in $g$ can be approximated to be linear.

This is represented by the following formula:

g' = g\left(1 - \frac{2h}{R}\right)

Here's how this formula is derived: The force of gravity ( $F$ ) is given by the universal law of gravitation:

F = G \frac{m_1 m_2}{d^2}

where: $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects (in this case, the Earth and the object we're considering), $d$ is the distance between the centers of the two objects.

If we consider an object of mass $m$ at the surface of Earth, the force it experiences due to gravity is:

F = G \frac{mM}{R^2} = mg

where: $M$ is the mass of Earth, $R$ is the radius of Earth, $g = \dfrac{GM}{R^2}$ is the acceleration due to gravity at the Earth's surface.

Now, if the object is at a height $h$ above the Earth's surface, the force it experiences is:

F' = G \frac{mM}{(R + h)^2}

We can write $(R + h)^2$ as $R^2 (1 + \dfrac{h}{R})^2$ , and because $h << R$ , we can use the binomial approximation $(1 + x)^2 \approx 1 + 2x$ for small $x$ to get:

F' \approx G \frac{mM}{R^2} \left(1 - \frac{2h}{R}\right) = mg \left(1 - \frac{2h}{R}\right)

Equating the forces $F = mg$ and $F' = mg'$ gives us:

g' = g \left(1 - \frac{2h}{R}\right)