Gravitation — Page 12 | JEE Physics

Q111

Two satellites of masses m and 3m revolve around the earth in circular orbits of radii r & 3r respectively. The ratio of orbital speeds of the satellites respectively is

A 3 : 1

B

\sqrt3

: 1

C 1 : 1

D 9 : 1

Correct Answer

Option B

Solution

The orbital speed of an object moving in a circular orbit around Earth (or any other celestial body) is given by the formula: $v = \sqrt{\dfrac{GM}{r}}$ where (v) is the orbital speed, (G) is the gravitational constant, (M) is the mass of the central body (Earth, in this case), and (r) is the radius of the orbit.

For the two satellites of masses (m) and (3m) in orbits of radii (r) and (3r) respectively, the ratio of their orbital speeds ( $v_1/v_2$ ) is: $\dfrac{v_1}{v_2} = \sqrt{\dfrac{\dfrac{GM}{r}}{\dfrac{GM}{3r}}} = \sqrt{\dfrac{3r}{r}} = \sqrt{3}$ So, the ratio of the orbital speeds of the satellites is ( $\sqrt{3} : 1$ ), which corresponds to Option B.

The inclusion of two different masses for the satellites, $m$ and $3m$ , in the problem might initially seem to suggest that the masses would influence their orbital speeds.

However, when it comes to circular orbital motion, especially around a much larger body like the Earth, the mass of the orbiting satellite does not directly affect its orbital speed.

This is because the orbital speed equation: $v = \sqrt{\dfrac{GM}{r}}$ only takes into account the mass of the central body (in this case, Earth's mass $M$ ), and the radius of the orbit $(r)$ , where $G$ is the gravitational constant.

Q112

The orbital angular momentum of a satellite is L, when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be -

A 9L

B 8L

C 4L

D 3L

Correct Answer

Option D

Solution

If we take the velocity (v) of the satellite to be $v = \sqrt{GM/r}$ , where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is: $L = mvr = m \sqrt{GMr}$ If the distance (r) from the Earth's center is increased by a factor of 8, the new angular momentum $L'$ is: $L' = m \sqrt{GM(8r)} = m \sqrt{8GMr} = 2 \sqrt{2} L$ However, the distance from the earth's surface is given, so we have to take into account the radius of the Earth ( $R$ ) in our calculations.

When the height from the earth's surface is increased eight times, the distance from the earth's center is $r = R + 8h$ , which is approximately $9R$ (because the height of the satellite above the Earth is generally much less than the radius of the Earth).

So, the new angular momentum (L'') is: $L'' = m \sqrt{GM(9R)} = 3 \sqrt{GM(R)} = 3L$

Q113

The weight of a body on the earth is

400 \mathrm{~N}

. Then weight of the body when taken to a depth half of the radius of the earth will be:

A 300 N

B 200 N

C 100 N

D Zero

Correct Answer

Option B

Solution

The gravitational field inside a uniform spherical body varies linearly with distance from the center.

If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.

The weight of an object at a depth $d$ from the Earth's surface is given by: $W_d = W_e (1 - \dfrac{d}{R})$ , where: $W_d$ is the weight at depth $d$ , $W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and $R$ is the radius of the Earth.

In this case, we're given that $W_e = 400 \, \text{N}$ , and we're asked to find the weight at a depth of $d = \dfrac{R}{2}$ .

Substituting these values into the formula, we get: $W_d = 400 \, \text{N} (1 - \dfrac{1}{2}) = 200 \, \text{N}$ .

Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.

Q114

Given below are two statements : one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

. Assertion A : Earth has atmosphere whereas moon doesn't have any atmosphere. Reason R : The escape velocity on moon is very small as compared to that on earth. 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 correct but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

C Both

\mathbf{A}

and

\mathbf{R}

are correct and

\mathbf{R}

is the correct explanation of

\mathbf{A}

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option C

Solution

The assertion (A) is true: Earth does have an atmosphere, while the Moon does not have a significant atmosphere.

The reason (R) is also true: The escape velocity on the Moon is indeed smaller than that on Earth.

The escape velocity is the minimum velocity an object must have to escape the gravitational pull of a planet or moon.

A smaller escape velocity means it's easier for particles (such as the particles that make up an atmosphere) to escape into space.

Additionally, the reason (R) is a correct explanation for the assertion (A).

The fact that the Moon's escape velocity is smaller than the Earth's is a major reason why the Moon doesn't have a significant atmosphere.

Over time, particles that could have made up an atmosphere have escaped the Moon's gravitational pull and dispersed into space.

Q115

A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing

\mathrm{W}

on earth will weigh on that planet:

A

2^{2 / 3} \mathrm{~W}

B W

C

2 \mathrm{~W}

D

2^{1 / 3} \mathrm{~W}

Correct Answer

Option D

Solution

The weight of an object on a planet is given by the equation $W = mg$ , where $m$ is the mass of the object and $g$ is the acceleration due to gravity.

The acceleration due to gravity on a planet is given by the equation $g = \dfrac{GM}{R^2}$ , where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

In this case, the mass of the planet is double that of Earth ( $M = 2M_E$ ), but the density is the same.

Density is defined as mass divided by volume, so if the mass is doubled and the density stays the same, the volume must also double.

Since the volume of a sphere (like a planet) is given by the equation $V = \dfrac{4}{3}\pi R^3$ , a doubling of the volume implies that the radius of the planet is $R = 2^{1/3}R_E$ .

Substituting these values back into the equation for $g$ , we get: $g_{\text{planet}} = G * \dfrac{2M_E}{(2^{1/3}R_E)^2} = 2^{1/3} * g_E$ So the weight of the object on the planet is $W_{\text{planet}} = m*g_{\text{planet}} = m2^{1/3} * g_E = 2^{1/3}W_E$

Q116

The weight of a body on the surface of the earth is

100 \mathrm{~N}

. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

A 50 N

B 64 N

C 25 N

D 100 N

Correct Answer

Option B

Solution

To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:

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

where

F

is the gravitational force,

G

is the gravitational constant,

m_1

and

m_2

are the masses of the two objects, and

r

is the distance between their centers. The weight of the body on the surface of the Earth is given as

100\,\text{N}

, which is also the gravitational force acting on it:

F_\text{surface} = G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}

When the body is taken to a height equal to one-fourth the radius of the Earth, the distance between the centers of the body and the Earth becomes

r_\text{new} = R_\text{earth} + \frac{1}{4} R_\text{earth} = \frac{5}{4} R_\text{earth}

. Now the gravitational force acting on the body at this new height is:

F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{r_\text{new}^2}

F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}

To find the ratio between the new gravitational force and the original force on the surface, we can write:

\frac{F_\text{new}}{F_\text{surface}} = \frac{G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}}{G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}}

Canceling out the common terms, we get:

\frac{F_\text{new}}{F_\text{surface}} = \frac{R_\text{earth}^2}{\left(\frac{5}{4} R_\text{earth}\right)^2} = \frac{1}{\left(\frac{5}{4}\right)^2} = \frac{1}{\left(\frac{25}{16}\right)} = \frac{16}{25}

Now, since the weight of the body on the surface is

100\,\text{N}

, we can find the new gravitational force as:

F_\text{new} = \frac{16}{25} \times 100\,\text{N} = 64\,\text{N}

So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is

64\,\text{N}

.

Q117

Choose the incorrect statement from the following:

A The linear speed of a planet revolving around the sun remains constant.

B When a body falls towards earth, the displacement of earth towards the body is negligible.

C The speed of satellite in a given circular orbit remains constant.

D For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant.

Correct Answer

Option A

Solution

The linear speed of a planet revolving around the sun does not remain constant, as planets follow an elliptical orbit around the sun.

According to Kepler's second law, the areal velocity of a planet remains constant, which means that the planet moves faster when it is closer to the sun (at perihelion) and slower when it is farther away (at aphelion).

Q118

A light planet is revolving around a massive star in a circular orbit of radius

\mathrm{R}

with a period of revolution T. If the force of attraction between planet and star is proportional to

\mathrm{R}^{-3 / 2}

then choose the correct option :

A

\mathrm{T}^2 \propto \mathrm{R}^{7 / 2}

B

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

C

\mathrm{T}^2 \propto \mathrm{R}^{5 / 2}

D

\mathrm{T}^2 \propto \mathrm{R}^{3 / 2}

Correct Answer

Option C

Solution

To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\mathrm{R}^{-3 / 2}$ .

According to Newton's law of universal gravitation, the force of attraction $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by

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

where $G$ is the gravitational constant.

However, in this particular case, the force of attraction is given to be proportional to $\mathrm{R}^{-3 / 2}$ , so we can write

F \propto \frac{1}{R^{3/2}}.

Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in orbit must be provided by this gravitational force.

Hence, we can write that

\frac{m v^2}{R} \propto \frac{1}{R^{3/2}},

where $m$ is the mass of the planet and $v$ is its orbital speed. Simplifying this, we get

v^2 \propto \frac{1}{R^{1/2}}.

Now, the speed $v$ can be related to the period T through the circumference of the orbit, which is given by $2\pi R$ .

The orbital speed is the circumference divided by the period:

v = \frac{2\pi R}{T}.

Substituting this into our proportionality, we get

\left( \frac{2\pi R}{T} \right)^2 \propto \frac{1}{R^{1/2}},

which simplifies to

\frac{4\pi^2 R^2}{T^2} \propto \frac{1}{R^{1/2}}.

Solving for $T^2$ , we get

T^2 \propto \frac{R^{2 + 1/2}}{4\pi^2},

so

T^2 \propto R^{5/2}.

Therefore, the correct option is Option C

T^2 \propto R^{5/2}.

Q119

If

\mathrm{G}

be the gravitational constant and

\mathrm{u}

be the energy density then which of the following quantity have the dimensions as that of the

\sqrt{\mathrm{uG}}

:

A Gravitational potential

B pressure gradient per unit mass

C Energy per unit mass

D Force per unit mass

Correct Answer

Option D

Solution

To determine the dimension of the quantity

\sqrt{uG}

, we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).

The gravitational constant

G

has dimensions given by:

[G] = M^{-1}L^{3}T^{-2}

where

M

stands for mass,

L

for length, and

T

for time. Energy density

u

is defined as the energy per unit volume. Since energy has dimensions of

ML^{2}T^{-2}

(from the dimension of work or energy, which is force times distance, and force itself has dimension

MLT^{-2}

), and volume has dimensions of

L^{3}

, the dimensions of energy density would be:

[u] = \frac{ML^{2}T^{-2}}{L^{3}} = M L^{-1} T^{-2}

Now, we find the dimensions of

\sqrt{uG}

by multiplying the dimensions of

u

and

G

, and then taking the square root:

[\sqrt{uG}] = \sqrt{[u][G]} = \sqrt{(M L^{-1} T^{-2})(M^{-1}L^{3}T^{-2})} = \sqrt{L^{2}T^{-4}} = LT^{-2}

So, the dimension of

\sqrt{uG}

is

LT^{-2}

, which corresponds to acceleration (length per square time).

Now, let's match this with the provided options: Option A (Gravitational potential) has dimensions of

[L^{2}T^{-2}]

, not matching our target of

LT^{-2}

. Option B (Pressure gradient per unit mass) would have dimensions of

[M^{-1}L^{-2}T^{-2}][L^{-1}]

(

Pressure\ Gradient = \frac{Pressure}{Length} = \frac{ML^{-1}T^{-2}}{L}

, and then divided by mass,

M

), which simplifies to

L^{-3}T^{-2}M^{-1}

, not matching. Option C (Energy per unit mass) has dimensions

ML^{2}T^{-2}M^{-1}

which simplifies to

L^{2}T^{-2}

, also not a match for the target dimension. Option D (Force per unit mass) has dimensions

MLT^{-2}M^{-1}

which simplifies directly to

LT^{-2}

, an exact match for our target dimension.

Thus, the correct answer is Option D (Force per unit mass), which has the same dimensions as that of

\sqrt{\mathrm{uG}}

.

Q120

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is : (Given

=

Radius of geo-stationary orbit for earth is

4.2 \times 10^4 \mathrm{~km}

)

A

1.68 \times 10^5 \mathrm{~km}

B

1.4 \times 10^4 \mathrm{~km}

C

8.4 \times 10^4 \mathrm{~km}

D

1.05 \times 10^4 \mathrm{~km}

Correct Answer

Option D

Solution

\begin{aligned} & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2} \sqrt{\frac{m_2}{m_1}} \\ & \Rightarrow \quad \frac{24}{6}=\left(\frac{4.2 \times 10^4}{r_2}\right)^{3 / 2} \sqrt{\frac{m / 4}{m}} \\ & \Rightarrow r_2=1.05 \times 10^4 \mathrm{~km} \end{aligned}