Gravitation — Page 7 | JEE Physics

Q61

The maximum and minimum distances of a comet from the Sun are 1.6

\times

1012 m and 8.0

\times

1010 m respectively. If the speed of the comet at the nearest point is 6

\times

104 ms

-

1, the speed at the farthest point is :

A 3.0

\times

103 m/s

B 6.0

\times

103 m/s

C 1.5

\times

103 m/s

D 4.5

\times

103 m/s

Correct Answer

Option A

Solution

v1 = 6 $\times$ 104 m/s Let point 1 is nearest point, and point 2 is farthest point.

Given, r1 = 8 $\times$ 1010 m & r2 = 1.6 $\times$ 1012 m By angular momentum conservation L1 = L2 mr1v1 = mr2v2 $\Rightarrow$ v2 =

{{{r_1}{v_1}} \over {{r_2}}}

$\therefore$ v2 =

{{8 \times {{10}^{10}} \times 6 \times {{10}^4}} \over {1.6 \times {{10}^{12}}}}

$\Rightarrow$ v2 = 3.0 $\times$ 103 m/s

Q62

The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

A 9 T

B 27 T

C 12 T

D 3 T

Correct Answer

Option B

Solution

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit.

This relationship can be written as:

T^2 \propto r^3

where: $T$ is the orbital period $r$ is the radius of the circular orbit (which serves as the semi-major axis in this case) Considering two satellites, one with period $T$ and radius $R$ , and another with unknown period $T'$ and radius $9R$ , we can form an equation:

\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}

This simplifies to:

\frac{{T'}^2}{T^2} = 729

Taking the square root of both sides, we get:

T' = T \times \sqrt{729} = T \times 27

So, the period of another satellite in a circular orbit of radius $9R$ is $27T$ .

Q63

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : [Take g = 10 ms

-

2, the radius of earth, R = 6400

\times

103 m, Take

\pi

= 3.14]

A 84 minutes

B 1200 minutes

C 60 minutes

D does not change

Correct Answer

Option A

Solution

For objects to float mg = 2 $\omega$ 2R $\omega$ = angular velocity of earth. R = Radius of earth

\omega = \sqrt {{g \over R}}

..... (1) Duration of day = T

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

..... (2)

\Rightarrow T = 2\pi \sqrt {{R \over g}}

= 2\pi \sqrt {{{6400 \times {{10}^3}} \over {10}}}

\Rightarrow {T \over {60}}

= 83.775 minutes

\simeq

84 minuites

Q64

The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is

{\overrightarrow L }

. The magnitude of the areal velocity of the planet is :

A

{{2L} \over M}

B

{{L} \over 2M}

C

{{L} \over M}

D

{{4L} \over M}

Correct Answer

Option B

Solution

Gravitational force line passes through the sun so torque about sun always zero for the planet.

$\therefore$ Angular momentum about sum is constant.

\overrightarrow L

= Constant we know, L = M

{v_ \bot }

r Now,

dA = {1 \over 2}

$\times$ base $\times$ height

= {1 \over 2} \times r \times {v_ \bot }dt

\Rightarrow {{dA} \over {dt}} = {{r{v_ \bot }} \over 2} = {r \over 2} \times {L \over {Mr}} = {L \over {2M}}

Q65

Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA an rB, respectively. If TA and TB are the time period of star A and star B, respectively, Then :

A

{{{T_A}} \over {{T_B}}} = {\left( {{{{r_A}} \over {{r_B}}}} \right)^{{3 \over 2}}}

B

{T_A} = {T_B}

C

{T_A} > {T_B}

(if

{m_A} > {m_B}

)

D

{T_A} > {T_B}

(if

{r_A} > {r_B}

)

Correct Answer

Option B

Solution

In a binary star system, the two stars orbit around a common center of mass.

When considering periods of revolution, Kepler's Third Law comes into play.

This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit.

It's often written in the following form for a single object orbiting another: T² ∝ r³ For a binary star system, this would still hold true.

The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii.

In other words, star A and star B complete one orbit in the same amount of time.

So, Option B: TA = TB is correct.

The other options (Option A, C, and D) would not be correct.

Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.

Q66

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ___________ s.

A

\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]

B

{1 \over 3}\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]

C

\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]

D

{1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]

Correct Answer

Option D

Solution

{1 \over 2}m{v^2} - {{GMm} \over r} = 0 \Rightarrow v = \sqrt {{{2GM} \over r}}

{{dr} \over {dt}} = \sqrt {{{2GM} \over r}}

\Rightarrow \int\limits_{{R_e}}^{({R_e} + h)} {\sqrt r dr = \int\limits_0^t {\sqrt {2GM} dt} }

\Rightarrow {2 \over 3}\left[ {{{({R_e} + h)}^{3/2}} - R_e^{3/2}} \right] = (t)\sqrt {2GM}

\Rightarrow t = {1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]

Q67

The minimum and maximum distances of a planet revolving around the sun are x1 and x2. If the minimum speed of the planet on its trajectory is v0 then its maximum speed will be :

A

{{{v_0}x_1^2} \over {x_2^2}}

B

{{{v_0}x_2^2} \over {x_1^2}}

C

{{{v_0}x_1^{}} \over {x_2^{}}}

D

{{{v_0}x_2^{}} \over {x_1^{}}}

Correct Answer

Option D

Solution

Angular momentum conservation equation

{v_0}{x_2} = {v_1}{x_1}

{v_1} = {{{v_0}{x_2}} \over {{x_1}}}

Q68

Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :

A

\sqrt {{G \over {2{R^3}}}}

B

{1 \over 2}\sqrt {{G \over {{R^3}}}}

C

{1 \over {2R}}\sqrt {{1 \over G}}

D

{{2G} \over {{R^3}}}

Correct Answer

Option B

Solution

The problem describes two identical particles of mass m=1kg each moving in a circle of radius R under the action of their mutual gravitational attraction.

This means that the particles are moving around a common center, and the distance between the particles is 2R (as the diameter of the circle).

In this case, the force providing the centripetal force for each particle to move in a circular path is the gravitational force between the particles.

The gravitational force between two masses m1 and m2 separated by a distance r is given by Newton's law of universal gravitation:

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

Since the two particles are identical, m1=m2=m=1kg.

And the distance between them r is 2R.

Substituting these into the above equation gives the gravitational force between the two particles:

F = G \frac{{m^2}}{{(2R)^2}} = G \frac{{1}}{{4R^2}}

This gravitational force is also equal to the centripetal force needed for each particle to move in a circular path of radius R.

The centripetal force is given by:

F = m R ω^2

Setting these two equations equal to each other gives:

G \frac{{1}}{{4R^2}} = 1 \cdot R \cdot ω^2

Rearranging this to solve for ω (the angular speed) gives:

ω = \frac{1}{2} \sqrt{\frac{G}{R^3}}

So, the angular speed of each particle is: $\dfrac{1}{2} \sqrt{\dfrac{G}{R^3}}$ .

Q69

The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0

\times

103 km. Find the mass of Mars.

\left\{ {Given\,{{4{\pi ^2}} \over G} = 6 \times {{10}^{11}}{N^{ - 1}}{m^{ - 2}}k{g^2}} \right\}

A 5.96

\times

1019 kg

B 3.25

\times

1021 kg

C 7.02

\times

1025 kg

D 6.00

\times

1023 kg

Correct Answer

Option D

Solution

Option D is correct.

{T^2} = {{4{\pi ^2}} \over {GM}}.{r^3}

M = {{4{\pi ^2}} \over G}.{{{r^3}} \over {{T^2}}}

by putting values

M = 6 \times {10^{23}}

Q70

Inside a uniform spherical shell : (1) the gravitational field is zero (2) the gravitational potential is zero (3) the gravitational field is same everywhere (4) the gravitational potential is same everywhere (5) all of the above Choose the most appropriate answer from the options given below :

A (1), (3) and (4) only

B (5) only

C (1), (2) and (3) only

D (2), (3) and (4) only

Correct Answer

Option A

Solution

Inside a spherical shell, gravitational field is zero and hence potential remains same everywhere.