Chapter 7: Gravitation

Ans: No.

The gravitational force of matter on nearby objects cannot be eliminated by any means. This is because gravitational force is independent of the nature of the medium's material.

Ans: Yes.

If the space station's size is large enough, then the astronaut will identify the change in Earth's gravity.

Ans: The tidal effect depends on the inverse of distance's cube while gravitational force depends on the inverse of distance's square. 

Since the distance between the Moon and the Earth is shorter than the distance between the Sun and the Earth, the tidal influence of the Moon's pull is higher than the tidal effect of the Sun's influence.

Ans: Acceleration due to gravity decreases with increasing altitude.

Acceleration due to gravity at height \[h\] is given by:

\[{{g}_{h}}=\left( 1-\frac{2h}{{{R}_{e}}} \right)g\]

Where, \[{{R}_{e}}=\]Radius of the Earth

\[g=\]Acceleration due to gravity on the surface of the Earth

It is clear from the relation that acceleration due to gravity lowers with a height increment.

Ans: Acceleration due to gravity decreases with increasing depth.

Acceleration due to gravity at depth \[d\] is given by:

\[{{g}_{d}}=\left( 1-\frac{d}{{{R}_{e}}} \right)g\]

It is clear from the relation that acceleration due to gravity lowers with a depth increment.

Ans: Acceleration due to gravity is independent of mass of the body.

Acceleration due to gravity of body having mass \[m\] is given by: \[g=\frac{GM}{{{R}^{2}}}\]

Where, \[G\]= Universal gravitational constant

\[M\]= Mass of the Earth

\[R\]= Radius of the Earth

Hence, it is clear that acceleration due to gravity is not dependent on the body's mass.

Ans: Gravitational potential energy of two points at \[{{r}_{2}}\] and \[{{r}_{1}}\] distance away from the Earth centre is given by:

\[V({{r}_{1}})=-\frac{GmM}{{{r}_{1}}}\] and

\[V({{r}_{2}})=-\frac{GmM}{{{r}_{2}}}\]

\[\therefore \] Difference in potential energy,

\[V=V({{r}_{2}})-V({{r}_{1}})\]

\[\Rightarrow \text{V}=-GMm\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\]

Hence, this formula is more reliable than the formula \[mg({{r}_{2}}-{{r}_{1}})\].

Ans: Time taken by the Earth to complete one revolution, \[{{T}_{e}}=1\text{ year}\].

Orbital radius of the Earth in its orbit, \[{{R}_{e}}=1AU\].

Time taken by the planet to complete one revolution, \[{{T}_{p}}=\frac{1}{2}{{T}_{e}}=\frac{1}{2}year\].

Orbital radius of the planet \[={{R}_{p}}\].

From Kepler's third law of planetary motion, we can write:

\[{{\left(\frac{{{R}_{p}}}{{{R}_{e}}} \right)}^{3}}={{\left( \frac{{{T}_{p}}}{{{T}_{e}}} \right)}^{2}}\]

\[\Rightarrow\frac{{{R}_{p}}}{{{R}_{e}}}={{\left(\frac{{{T}_{p}}}{{{T}_{e}}} \right)}^{\frac{2}{3}}}\]

\[\Rightarrow\frac{{{R}_{p}}}{{{R}_{e}}}={{\left(\frac{\frac{1}{2}}{1} \right)}^{\frac{2}{3}}}\]

\[\Rightarrow \frac{{{R}_{p}}}{{{R}_{e}}}={{(0.5)}^{\frac{2}{3}}}\]

\[\Rightarrow \frac{{{R}_{p}}}{{{R}_{e}}}=0.63\]

Hence, the orbital radius of the planet will be \[0.63\] times smaller than that of the Earth.

Ans: Given that,

Orbital period of \[{{I}_{O}}={{T}_{{{I}_{O}}}}=1.769days=1.769\times 24\times 60\times 60s\]

Orbital radius of \[{{I}_{O}}={{R}_{{{I}_{O}}}}=4.22\times {{10}^{8}}m\]

Satellite \[{{I}_{O}}\] is revolving around Jupiter.

Mass of the Jupiter is given by:

\[{{M}_{J}}=\frac{4{{\pi }^{2}}R_{{{I}_{o}}}^{3}}{GT_{{{I}_{o}}}^{2}}\]......(i)

Where, \[{{M}_{J}}\]= Mass of Jupiter

\[G\]is the Universal gravitational constant

Orbital period of the Earth,

\[{{T}_{e}}=365.25days=365.25\times 25\times 60\times 60s\]  

Orbital radius of the Earth,

\[{{R}_{e}}=1AU=1.496\times {{10}^{11}}m\]  

Mass of sun is given as:

\[{{M}_{s}}=\frac{4{{\pi }^{2}}R_{e}^{3}}{GT_{e}^{2}}\]......(ii)

\[\Rightarrow\frac{{{M}_{S}}}{{{M}_{J}}}=\frac{4{{\pi}^{2}}R_{e}^{3}}{GT_{e}^{2}}\times\frac{GT_{Io}^{2}}{4{{\pi}^{2}}R_{Io}^{3}}=\frac{R_{e}^{3}}{R_{Io}^{3}}\times \frac{T_{Io}^{2}}{T_{e}^{2}}\]

\[\Rightarrow \frac{{{M}_{S}}}{{{M}_{J}}}={{\left( \frac{1.769\times 24\times 60\times 60}{365.25\times 24\times 60\times 60} \right)}^{2}}\times {{\left( \frac{1.496\times {{10}^{11}}}{4.22\times {{10}^{8}}} \right)}^{3}}\]

\[\Rightarrow \frac{{{M}_{S}}}{{{M}_{J}}}=1045.04\]

\[\therefore \frac{{{M}_{s}}}{{{M}_{J}}}\sim 1000\]

We get,

\[{{M}_{s}}\sim 1000\times {{M}_{J}}\]  

Hence, it can be concluded that the mass of Jupiter is about one-thousandth that of the Sun.

Ans: Mass of galaxy Milky way, \[M=2.5\times {{10}^{11}}\] solar mass

Solar mass = Mass of sun\[=2.0\times {{10}^{36}}kg\]

Mass of our galaxy, \[M=2.5\times {{10}^{11}}\times 2\times {{10}^{35}}\]

\[M=5\times {{10}^{41}}kg\]

Diameter of Milky Way, \[d={{10}^{5}}ly\]

Radius of Milky Way, \[r=5\times {{10}^{4}}ly\]

\[1\text{ }ly=9.46\times 1{{0}^{15}}m\]

\[r=5\times {{10}^{4}}\times 9.46\times 1{{0}^{15}}\]

\[\Rightarrow \text{r}=4.73\times {{10}^{20}}m\]

Since a star rotates around the galactic centre of the Milky Way, its time period is given by: 

\[T={{\left( \frac{4{{\pi }^{2}}{{r}^{3}}}{GM} \right)}^{\frac{1}{2}}}=\left( \frac{4\times {{\left( 3.14 \right)}^{2}}\times {{\left( 4.73 \right)}^{3}}\times {{10}^{60}}}{6.67\times {{10}^{-11}}\times 5\times {{10}^{41}}} \right)\]

\[\Rightarrow T={{\left( 125.27\times {{10}^{30}} \right)}^{\frac{1}{2}}}\]

\[\Rightarrow \text{T}=1.12\times {{10}^{16}}s\]

As we know,

\[1\text{ }year=365\times 24\times 60\times 60s\]

We get,

\[1s=\frac{1}{365\times 24\times 60\times 60}years\]

\[\Rightarrow 1.12\times {{10}^{16}}s=\frac{1.12\times {{10}^{16}}}{365\times 24\times 60\times 60}\]

\[\therefore 1.12\times {{10}^{16}}s=3.55\times {{10}^{8}}years\]

Star will take  \[3.55\times {{10}^{8}}years\] to complete one revolution.

Ans: Kinetic energy.

The sum of kinetic energy and potential energy is the total mechanical energy. The satellite's gravitational potential energy is zero at infinity. As we know, the Earth-satellite system is a bound system; then, the satellite's total energy is negative.

Thus, at infinity, the orbiting satellite's total energy is equal to the negative of its kinetic energy. 

Ans: Less.

An orbiting satellite gets a certain amount of energy that permits it to rotate around the Earth. Its orbit gives this energy. It requires comparatively lesser energy to go out of the influence of the Earth's gravitational field than a fixed object on the Earth's surface that initially holds no energy.

Ans: No.

Escape velocity of a body from the Earth is given by the relation:

\[v=\sqrt{2gR}\]

The escape speed of a body does not depend on the mass of the body.

Ans: No.

Escape velocity of a body from the Earth is given by the relation:

\[v=\sqrt{2gR}\]

The escape speed of a body does not depend on the location from where it is projected.

Ans: No.

Escape velocity of a body from the Earth is given by the relation:

\[v=\sqrt{2gR}\]

The escape speed of a body does not depend on the direction of projection.

Ans: Yes.

Escape velocity of a body from the Earth is given by the relation:

\[v=\sqrt{2gR}\]

The escape speed of a body does slightly depend on the height of the location from where the body is launched. This is because escape velocity is slightly dependent on the gravitational potential at a certain height. 

Ans: No.

The expression for angular momentum is:

$L=\frac{m{{v}^{2}}}{r}$

As angular momentum is constant, linear speed varies as $r$ varies.

Ans: No.

The expression for angular momentum is:

$L=mw{}^{2}r$

As angular momentum is constant, angular speed varies as  $r$ varies.

Ans: Yes.

There is no external torque and thus, the angular momentum is constant. 

Ans: No.

The expression for kinetic energy is:

$K=\frac{1}{2}m{{v}^{2}}$

Kinetic energy is not constant because linear speed varies.

E. Potential Energy?

Ans: No.

By law of conservation of energy, total energy is constant and it is the sum of potential energy and kinetic energy. 

As kinetic energy is not constant, potential energy is also not constant.

F. Total Energy Throughout Its Orbit? Neglect Any Mass Loss of the Comet When it Comes Very Close to the Sun.

Ans: Yes.

By law of conservation of energy, the total energy is constant.

Ans: Legs hold the whole mass in a standing place due to gravitational force. 

In space, an astronaut appears weightless because of the absence of gravity. Therefore, swollen feet of a spaceman do not affect him/her in space.

Ans: A swollen face is usually caused because of seeming weightlessness in space. 

Sense organs such as the nose, eyes, ears, and mouth establish a person's face. This symptom can affect a spacewalker in space.

Ans: Headaches are because of mental stress. 

It can influence the working of an astronaut in space.

Ans: Space has diverse orientations. 

Therefore, orientational difficulty can affect an astronaut in space.

Ans: Option (iii) is correct.

Gravitational potential (\[V\]) is fixed at all points in a spherical shell. Hence, the potential gravitational gradient \[\left( \frac{dV}{dr} \right)\] is zero everywhere inside the spherical surface. 

The potential gravitational gradient is equivalent to the negative of gravitational intensity. Hence, intensity is also zero at all locations inside the spherical shell. This shows that gravitational forces operating at a point in a spherical shell are symmetric.

If the top half of a spherical shell is cut out, then the net gravitational force working on a particle located at centre O will be downward. 

Since gravitational intensity is described as the gravitational force per unit mass at that location, it will also act downward. Thus, the gravitational intensity at the centre of the given hemispherical shell has the direction indicated by the arrow \[c\].