Gravitation — Page 14 | JEE Physics

Q131

A

90 \mathrm{~kg}

body placed at

2 \mathrm{R}

distance from surface of earth experiences gravitational pull of : (

\mathrm{R}=

Radius of earth,

\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}

)

A 300 N

B 225 N

C 100 N

D 120 N

Correct Answer

Option C

Solution

The gravitational force

F

that an object of mass

m

placed at a distance

r

(from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:

F = \frac{G M m}{r^2}

Where:

G

is the gravitational constant (

6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2

)

M

is the mass of the Earth (approximately

5.972 \times 10^{24} \text{kg}

)

m

is the mass of the object

r

is the distance from the center of the Earth to the object However, when the object is at a distance

2R

from the surface of the Earth, the total distance from the center of the Earth

r'

becomes

R + 2R = 3R

. This is because the radius of the Earth

R

is the distance from the Earth's center to its surface, so if the object is

2R

above the surface, the total distance from the center is

3R

. At the surface of the Earth, the gravitational force (

F_{earth}

) that acts on an object is given by its weight, which can be calculated using the formula

F_{earth} = m \cdot g

, where

g

is the acceleration due to gravity on the surface of the Earth. Given that

g = 10 \text{m/s}^2

and the mass of the body

m = 90 \text{kg}

, we get:

F_{earth} = 90 \text{kg} \cdot 10 \text{m/s}^2 = 900 \text{N}

To find the gravitational pull at a distance

2R

from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth.

Since the distance triples (

3R

from

R

), the gravitational force becomes

\frac{1}{3^2} = \frac{1}{9}

of the force at the surface. Therefore, the gravitational pull

F_{2R}

on the body when placed at

2R

from the Earth's surface is:

F_{2R} = \frac{F_{earth}}{9} = \frac{900 \text{N}}{9} = 100 \text{N}

Thus, the gravitational pull on the body when it is placed at a distance of

2R

from the Earth's surface is 100 N. So, the correct option is: Option C: 100 N

Q132

Correct formula for height of a satellite from earths surface is :

A

\left(\dfrac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R

B

\left(\dfrac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R

C

\left(\dfrac{T^2 R^2 g}{4 \pi^2}\right)^{-1 / 3}+R

D

\left(\dfrac{T^2 R^2}{4 \pi^2 g}\right)^{1 / 3}-R

Correct Answer

Option A

Solution

\begin{aligned} & T=2 \pi \sqrt{\frac{r^3}{G M}} \\ & T^2=\frac{4 \pi^2}{G M}(R+h)^3 \\ & h=\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{G M \cdot R}{R^2} \cdot \frac{T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{\frac{1}{3}}-R \end{aligned}

Q133

Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of

\mathrm{A}

is

3 v

, the speed of

\mathrm{B}

will be :

A

6 v

B

\dfrac{4}{3} v

C

3 v

D

12 v

Correct Answer

Option A

Solution

To solve this question, we will use the fact that for an object in a circular orbit, the centripetal force required to keep the object in orbit is provided by the gravitational force between the object and the planet it is orbiting.

This principle gives us the relationship between the speed of the satellite, its orbital radius, and the mass of the planet.

The formula for the gravitational force is given by:

F = \frac{G M m}{r^2}

where: $F$ is the gravitational force between the two masses, $G$ is the gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the satellite, and $r$ is the distance between the center of the planet and the satellite (radius of the orbit).

The centripetal force required to keep the satellite in orbit is given by:

F_c = \frac{m v^2}{r}

where: $F_c$ is the centripetal force, $v$ is the orbital speed of the satellite, and $r$ is the radius of the orbit.

Since the gravitational force is providing the centripetal force, we can set the two forces equal to each other to find the relationship between speed and radius:

\frac{G M m}{r^2} = \frac{m v^2}{r}

By simplifying this, we find the equation for the orbital speed of a satellite:

v = \sqrt{\frac{G M}{r}}

Now, we can compare the speeds of satellites $A$ and $B$ based on their radii.

Satellite $A$ orbits at a radius of $4R$ with a speed of $3v$ , and we need to find the speed of satellite $B$ which orbits at a radius of $R$ .

Substituting the respective radii into the speed equation: For satellite $A$ , $v_A = 3v = \sqrt{\dfrac{G M}{4R}}$ For satellite $B$ , we want to find $v_B$ given that its radius is $R$ : $v_B = \sqrt{\dfrac{G M}{R}}$ To find the ratio of $v_B$ to $3v$ (or the speed of $A$ ), we can write:

v_B = \sqrt{\frac{G M}{R}} = \sqrt{4} \sqrt{\frac{G M}{4R}} = 2 \cdot \sqrt{\frac{G M}{4R}}

Given that $\sqrt{\dfrac{G M}{4R}}$ is the speed of satellite $A$ divided by 3 ( $3v$ ), we find that:

v_B = 2 \cdot 3v = 6v

Therefore, the correct answer is Option A: $6 v$ .

Q134

To project a body of mass

m

from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is

R_E, g=

acceleration due to gravity on the surface of earth):

A

1 / 2 m g R_E

B

4 m g R_E

C

m g R_E

D

2 m g R_E

Correct Answer

Option C

Solution

The kinetic energy required to project a body of mass $m$ from the Earth's surface to infinity, also known as the escape kinetic energy, can be calculated using the concept of gravitational potential energy.

The escape velocity $v_e$ is the velocity a body must have to escape the gravitational field of the Earth without any further propulsion.

The formula for escape velocity is: $v_e = \sqrt{2gR_E}$ Where $g$ is the acceleration due to gravity on the surface of Earth and $R_E$ is the radius of the Earth.

The kinetic energy $K$ required for this is given by: $K = \dfrac{1}{2}mv_e^2$ Substituting the escape velocity formula into the kinetic energy formula gives: $K = \dfrac{1}{2}m\left(2gR_E\right)$ $K = \dfrac{1}{2} \times 2 \times mgR_E$ $K = mgR_E$ Therefore, the required kinetic energy to project a body of mass $m$ from Earth's surface to infinity is $mgR_E$ .

So, the correct answer is: Option C: $mgR_E$

Q135

If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon

=27

days and gravitational attraction between the satellite and the moon is neglected.

A 3 days

B 27 days

C 81 days

D 1 day

Correct Answer

Option D

Solution

To find the time period of the satellite, we can use Kepler’s third law, which states that the square of the orbital period is proportional to the cube of the orbital radius.

Mathematically, this is expressed as:

\frac{T^2}{R^3} = \text{constant}

Here’s how we solve the problem step by step: Let: The Moon's distance from Earth be

R_{\text{moon}}

. The Moon's orbital period be

T_{\text{moon}} = 27

days. The satellite's orbital radius be

R_{\text{sat}} = \frac{R_{\text{moon}}}{9}

(since it is 9 times closer). According to Kepler's law for both orbits, we have:

\frac{T_{\text{sat}}^2}{R_{\text{sat}}^3} = \frac{T_{\text{moon}}^2}{R_{\text{moon}}^3}

Substitute

R_{\text{sat}}

:

\frac{T_{\text{sat}}^2}{\left(\frac{R_{\text{moon}}}{9}\right)^3} = \frac{27^2}{R_{\text{moon}}^3}

Simplify the left side by noting that:

\left(\frac{R_{\text{moon}}}{9}\right)^3 = \frac{R_{\text{moon}}^3}{9^3} = \frac{R_{\text{moon}}^3}{729}

So the equation becomes:

\frac{T_{\text{sat}}^2}{\frac{R_{\text{moon}}^3}{729}} = \frac{27^2}{R_{\text{moon}}^3}

Multiply both sides by

\frac{R_{\text{moon}}^3}{729}

:

T_{\text{sat}}^2 = 27^2 \times \frac{1}{729}

Recognize that

27^2 = 729

:

T_{\text{sat}}^2 = \frac{729}{729} = 1

Take the square root:

T_{\text{sat}} = 1 \text{ day}

Thus, the satellite’s orbital period is 1 day, which corresponds to Option D: 1 day.

Q136

A satellite is launched into a circular orbit of radius ' R ' around the earth. A second satellite is launched into an orbit of radius 1.03 R . The time period of revolution of the second satellite is larger than the first one approximately by

A

3 \%

B

2.5 \%

C

4.5 \%

D

9 \%

Correct Answer

Option C

Solution

T \propto R^{\frac{3}{2}}

For a satellite in a circular orbit with radius

R

, the time period is given by

T = 2\pi \sqrt{\frac{R^3}{GM}}.

If the radius is increased to

1.03R

, the new time period

T'

becomes

T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}.

For a small change, we can approximate

(1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045.

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.

Q137

An object is kept at rest at a distance of

3 R

above the earth's surface where

R

is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is : (Assume

\mathrm{M}=

mass of earth,

\mathrm{G}=

Universal gravitational constant)

A

\sqrt{\dfrac{3 \mathrm{GM}}{\mathrm{R}}}

B

\sqrt{\dfrac{2 \mathrm{GM}}{\mathrm{R}}}

C

\sqrt{\dfrac{\mathrm{GM}}{2 \mathrm{R}}}

D

\sqrt{\dfrac{\mathrm{GM}}{\mathrm{R}}}

Correct Answer

Option C

Solution

\begin{aligned} &\begin{aligned} & \mathrm{P}_{\mathrm{P}}+\mathrm{k}_{\mathrm{P}}=\mathrm{P}_{\mathrm{o}}+\mathrm{k}_0 \\ & -\frac{\mathrm{GMm}}{4 \mathrm{R}}+\frac{1}{2} \mathrm{mV}_{\mathrm{P}}^2=0 \\ & \mathrm{~V}_{\mathrm{P}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \end{aligned}\\ &\text{ Choice } 1 \end{aligned}

Q138

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason

\mathbf{R}

Assertion A : The kinetic energy needed to project a body of mass

m

from earth surface to infinity is

\dfrac{1}{2} \mathrm{mgR}

, where R is the radius of earth. Reason R : The maximum potential energy of a body is zero when it is projected to infinity from earth surface. In the light of the above statements, choose the correct answer from the options given below

A

\mathbf{A}

is false but

\mathbf{R}

is true

B Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

C

\mathbf{A}

is true but

\mathbf{R}

is false

D Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Correct Answer

Option A

Solution

Let's analyze both the Assertion and the Reason step by step.

Calculation of the Required Kinetic Energy: The gravitational potential energy of a body of mass

m

at a distance

r

from the center of the Earth is given by:

U(r) = -\frac{GMm}{r},

where

G

is the gravitational constant and

M

is the mass of the Earth. At the Earth's surface (where

r = R

), the potential energy is:

U(R) = -\frac{GMm}{R}.

At infinity, the potential energy is defined as:

U(\infty) = 0.

The energy needed to move the body from the Earth’s surface to infinity is the change in potential energy:

\Delta U = U(\infty) - U(R) = 0 - \Bigg(-\frac{GMm}{R}\Bigg) = \frac{GMm}{R}.

Using the relation

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

we find:

\frac{GMm}{R} = mgR.

So, the required kinetic energy to just reach infinity (with zero speed at infinity) is:

\text{Kinetic Energy} = mgR.

The Assertion A states that the required kinetic energy is

\frac{1}{2}mgR

, which is only half of the actual value.

Verification of the Reason: The Reason states: "The maximum potential energy of a body is zero when it is projected to infinity from earth surface."

By convention in gravitational problems, the potential energy at infinity is taken to be zero.

This is a standard and correct statement.

Conclusion: Assertion A is false because it underestimates the needed energy; the correct kinetic energy should be

mgR

.

Reason R is true since the gravitational potential energy is indeed defined to be zero at infinity.

Therefore, the correct answer is: Option A:

\mathbf{A}

is false but

\mathbf{R}

is true.

Q139

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :

A 2W

B W

C

{2^{{1 \over 3}}}

W

D

\sqrt 2

W

Correct Answer

Option C

Solution

Density is same

M = {4 \over 3}\pi {R^3}\rho ,2m = {4 \over 3}\pi R{'^3}\rho

R' = {2^{1/3}}R

\omega = {{GMm} \over {{R^2}}}

{\omega _2} = {{G2Mm} \over {R{'^2}}}

{\omega _2} = {2^{1/3}}\omega

Q140

Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is ;

A 2

B

{{1 \over 2}}

C

\sqrt {{1 \over 2}}

D 1

Correct Answer

Option D

Solution

Orbital velocity V =

\sqrt {{{GMe} \over r}}

TA =

{1 \over 2}

mA V

_A^2

TB =

{1 \over 2}

mB V

_B^2

$\Rightarrow$

{{{T_A}} \over {{T_B}}} = {{m \times {{Gm} \over R}} \over {2m \times {{Gm} \over {2R}}}}

$\Rightarrow$

{{{T_A}} \over {{T_B}}}

= 1