Chapter 7: Gravitation

Ans: Option (ii) 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 P 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 arbitrary point P of the given hemispherical shell has the direction indicated by the arrow \[e\].

Ans: Mass of the Sun, \[{{M}_{s}}=2\times {{10}^{30}}kg\]

Mass of the Earth, \[{{M}_{e}}=6\times {{10}^{24}}kg\]

Orbital radius, \[r=1.5\times {{10}^{11}}m\]

Mass of the rocket=\[m\]

Let \[x\] be the distance from the Earth centre where the gravitational force working on the satellite \[P\] becomes zero.

From Newton's law of gravitation, we can equalize gravitational forces acting on satellite \[P\] under the effect of the Sun and the Earth as:

\[\frac{Gm{{M}_{s}}}{{{\left( r-x \right)}^{2}}}=Gm\frac{{{M}_{e}}}{{{x}^{2}}}\]

We get,

\[{{\left( \frac{r-x}{x} \right)}^{2}}=\frac{{{M}_{s}}}{{{M}_{e}}}\]

\[\Rightarrow \frac{r-x}{x}={{\left( \frac{2\times 1{{0}^{30}}}{60\times {{10}^{24}}} \right)}^{\frac{1}{2}}}=577.35\]

\[\Rightarrow 1.5\times {{10}^{11}}-x=577.35x\]

\[\Rightarrow 578.35x=1.5\times {{10}^{11}}\]

\[\Rightarrow x=\frac{1.5\times {{10}^{11}}}{578.35}=2.59\times {{10}^{8}}m\]

Therefore, \[2.59\times {{10}^{8}}m\] is the distance from the earth's centre at which the gravitational force on the rocket is zero.

Ans: Orbital radius of the Earth around the Sun, \[r=1.5\times {{10}^{11}}m\].

Time taken by the Earth to cover one revolution around the Sun,

\[T=1\text{ year}\]

\[\Rightarrow \text{T=}365.25\text{ days}\]

We get,

\[\text{T=365}\text{.25}\times \text{24}\times \text{60}\times \text{60}s\]

Universal gravitational constant, 

\[G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\]. 

Thus, we can use the relation given below for calculating the mass of the Sun,

\[M=\frac{4{{\pi }^{2}}{{r}^{3}}}{G{{T}^{2}}}\]

\[\Rightarrow \text{M}=\frac{4\times {{(3.14)}^{2}}\times {{(1.5\times {{10}^{11}})}^{3}}}{6.67\times {{10}^{-11}}\times {{(365.25\times 24\times 60\times 60)}^{2}}}\]

\[\Rightarrow \text{M}=\frac{133.24\times {{10}^{33}}}{6.64\times {{10}^{4}}}\]

\[\Rightarrow \text{M}=2.0\times {{10}^{30}}kg\]

Hence, the mass of the Sun is \[2\times {{10}^{30}}kg\].

Ans: Distance of the Earth from the Sun, \[{{r}_{e}}=1.5\times 1{{0}^{8}}km=1.5\times {{10}^{11}}m\]

Time period of the Earth \[={{T}_{e}}\]

Time period of Saturn, \[{{T}_{s}}=29.5{{T}_{e}}\]

Distance of Saturn from the Sun \[={{r}_{s}}\] 

From Kepler's third law of planetary motion, we have, 

\[T={{\left( \frac{4{{\pi }^{2}}{{r}^{3}}}{GM} \right)}^{\frac{1}{2}}}\] 

For Saturn and Sun, we can write,

\[\frac{{{r}_{s}}^{3}}{{{r}_{e}}^{3}}=\frac{{{T}_{s}}^{2}}{{{T}_{e}}^{2}}\]

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

\[\Rightarrow {{r}_{s}}=1.5\times {{10}^{11}}{{\left( \frac{29.5{{T}_{e}}}{{{T}_{e}}} \right)}^{\frac{2}{3}}}\]

\[\Rightarrow {{r}_{s}}=1.5\times {{10}^{11}}{{(29.5)}^{\frac{2}{3}}}\]

\[\Rightarrow {{r}_{s}}=1.5\times {{10}^{11}}\times 9.55\]

\[\Rightarrow {{r}_{s}}=14.32\times {{10}^{11}}m\]

Hence, the distance between the Sun and Saturn is  \[1.43\times {{10}^{12}}m\].

Ans: Weight of the body, \[W=63N\]

Acceleration due to gravity at height \[h\] from the Earth’s surface

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

For \[h=\frac{{{R}_{e}}}{2}\],  gravity at $h$ is given by:

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

\[\Rightarrow {{g}_{h}}=\frac{g}{{{\left( 1+\frac{1}{2} \right)}^{2}}}\]

\[\Rightarrow {{g}_{h}}=\frac{4}{9}g\]  

Weight of a body of mass \[m\] at height \[h\] is given as:

\[W'=m{{g}_{h}}\]

\[\Rightarrow W'=m\times \frac{4}{9}g\]

\[\Rightarrow W'=\frac{4}{9}\times mg\]

\[\Rightarrow \text{W }\!\!'\!\!\text{ }=\frac{4}{9}W\]

\[\Rightarrow \text{W }\!\!'\!\!\text{ }=\frac{4}{9}\times 63\]

\[\Rightarrow W'=28N\]

Thus, the gravitational force on the body due to the earth at a height equal to half the radius of the earth is $28N$.

Ans: Weight of a body of mass \[m\] at the Earth's surface is given by, 

\[W=mg=250N\]

Body of mass \[m\] is located at depth, \[d=\frac{1}{2}{{R}_{e}}\]

Where, \[{{R}_{e}}\]  is the radius of the Earth.

Acceleration due to gravity at depth \[{{g}_{d}}\] is given by the relation:

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

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

Weight of the body at depth \[d\],

\[W'=mg\]

\[\Rightarrow W'=m\times \frac{1}{2}g=\frac{1}{2}mg=\frac{1}{2}W\]

\[\Rightarrow W'=\frac{1}{2}\times 250=125N\]

The weight of body half way down to the centre of the earth is $125N.$

Ans: Distance from the centre of the Earth \[=8\times {{10}^{6}}m\]

Velocity of the rocket, \[v=5km{{s}^{-1}}=5\times {{10}^{3}}m{{s}^{-1}}\]

Mass of the Earth, \[{{M}_{e}}=6.0\times {{10}^{24}}kg\]

Radius of the Earth, \[{{R}_{e}}=6.4\times {{10}^{6}}m\]

Height reached by rocket mass $m$, is \[h\].

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy 

\[=\frac{1}{2}m{{v}^{2}}+\left( \frac{-G{{M}_{e}}m}{{{R}_{e}}} \right)\]

At highest point $h,$ \[v=0\]

And, Potential energy \[=-\frac{G{{M}_{e}}m}{{{R}_{e}}+h}\]

Total energy of the rocket\[=0+\left( \frac{G{{M}_{e}}m}{{{R}_{e}}+h} \right)=-\frac{G{{M}_{e}}m}{{{R}_{e}}+h}\]

From the law of energy conservation, we have

Total energy of the rocket at the Earth's surface = Total energy of rocket at height \[h\].

We have,

\[\frac{1}{2}m{{v}^{2}}+\left( -\frac{G{{M}_{e}}m}{{{R}_{e}}} \right)=-\frac{G{{M}_{e}}m}{{{R}_{e}}+h}\]

\[\Rightarrow \frac{1}{2}{{v}^{2}}=G{{M}_{e}}\left( \frac{1}{{{R}_{e}}}-\frac{1}{{{R}_{e}}+h} \right)\]

\[\Rightarrow \frac{1}{2}{{v}^{2}}=G{{M}_{e}}\left( \frac{{{R}_{e}}+h-{{R}_{e}}}{{{R}_{e}}({{R}_{e}}+h)} \right)\]

\[\Rightarrow \frac{1}{2}{{v}^{2}}=\frac{G{{M}_{e}}h}{{{R}_{e}}({{R}_{e}}+h)}\times \frac{{{R}_{e}}}{{{R}_{e}}}\]

\[\Rightarrow \frac{1}{2}{{v}^{2}}=\frac{g{{R}_{e}}h}{{{R}_{e}}+h}\]

Where, \[g=\frac{GM}{R_{e}^{2}}=9.8m{{s}^{-2}}\] , is the acceleration due to gravity on the Earth’s surface.

Clearly,

\[{{v}^{2}}({{R}_{e}}+h)=2g{{R}_{e}}h\]

\[\Rightarrow {{v}^{2}}{{R}_{e}}=h(2g{{R}_{e}}-{{v}^{2}})\]

\[\Rightarrow h=\frac{{{R}_{e}}{{v}^{2}}}{2g{{R}_{e}}-{{v}^{2}}}\]

\[\Rightarrow \text{h}=\frac{6.4\times 25\times {{10}^{12}}}{100.44\times 1{{0}^{6}}}\]

\[\Rightarrow \text{h}=1.6\times {{10}^{6}}m\]

Height achieved by the rocket with respect to the centre of the Earth, $H$ is given by: \[H={{R}_{e}}+h\]

\[\Rightarrow H=6.4\times {{10}^{6}}+1.6\times {{10}^{6}}\]

We get,

\[H=8.0\times {{10}^{6}}m\]

The distance from the earth is \[8.0\times {{10}^{6}}m\] where the rocket goes before returning to the earth.

Ans: Projectile’s escape velocity from the Earth, \[{{v}_{esc}}=11.2km{{s}^{-1}}\]

Projection velocity of the projectile, \[{{v}_{p}}={{3}_{{{v}_{esc}}}}\]

Mass of the projectile \[=m\]

Projectile’s velocity far away from the Earth \[={{v}_{f}}\]

Projectile’s total energy on the Earth  \[=\frac{1}{2}m{{v}_{p}}^{2}-\frac{1}{2}m{{v}_{esc}}^{2}\]

The projectile's gravitational potential energy far away from the Earth is zero.

Total energy of the projectile far away from the Earth \[=\frac{1}{2}m{{v}_{f}}^{2}\]

From the law of energy conservation, we have

\[\frac{1}{2}m{{v}_{p}}^{2}-\frac{1}{2}m{{v}_{esc}}^{2}=\frac{1}{2}m{{v}_{f}}^{2}\]  

We get,

\[{{v}_{f}}=\sqrt{{{v}_{p}}^{2}-{{v}_{esc}}^{2}}\]

\[\Rightarrow {{v}_{f}}=\sqrt{{{(3{{v}_{esc}})}^{2}}-{{({{v}_{esc}})}^{2}}}\]

\[\Rightarrow {{v}_{f}}=\sqrt{8}{{v}_{esc}}\]

\[\Rightarrow {{v}_{f}}=\sqrt{8}\times 11.2\]

\[\Rightarrow {{v}_{f}}=31.68km{{s}^{-1}}\]

Thus, the speed of the body far away from the earth is \[31.68km{{s}^{-1}}\].

Ans: Given that,

Mass of the Earth, \[{{M}_{e}}=6.0\times {{10}^{24}}kg\]

Mass of the satellite, \[m=200kg\]

Radius of the Earth, \[{{R}_{e}}=6.4\times {{10}^{6}}m\]

Universal gravitational constant, \[G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\]

Height of the satellite, \[h=400km=4\times {{10}^{5}}m=0.4\times {{10}^{6}}m\]

Total energy of the satellite at height $h$,

 \[TE=\frac{1}{2}m{{v}^{2}}+\left( \frac{-G{{M}_{e}}m}{{{R}_{e}}+h} \right)\]

Orbital velocity of the satellite, \[v=\sqrt{\frac{G{{M}_{e}}}{{{R}_{e}}+h}}\]

Total energy at height $h$, \[TE=\frac{1}{2}m\left( \frac{G{{M}_{e}}}{{{R}_{e}}+h} \right)-\frac{G{{M}_{e}}m}{{{R}_{e}}+h}=-\frac{1}{2}\left( \frac{G{{M}_{e}}m}{{{R}_{e}}+h} \right)\]

The negative sign explains that the satellite is attached to the Earth. This is called the bound energy of the satellite.

The energy needed to send the satellite out of its orbit is equal to the negative of Bound energy.

Bound energy is, \[BE=\frac{1}{2}\left( \frac{G{{M}_{e}}m}{{{R}_{e}}+h} \right)\]

Where, ${{M}_{e}}$ is the mass of the Earth.

${{R}_{e}}$ is the radius of the Earth.

h is the height.

m is the mass of the satellite.

\[\Rightarrow BE=\frac{1}{2}\times \frac{6.67\times {{10}^{-11}}\times 6.0\times {{10}^{24}}\times 200}{\left( 6.4\times {{10}^{6}}+0.4\times {{10}^{6}} \right)}\]

\[\Rightarrow BE=\frac{1}{2}\times \frac{6.67\times 6\times 2\times 10}{6.4\times {{10}^{6}}}\]

\[\Rightarrow BE=5.9\times {{10}^{9}}J\] 

Clearly, \[5.9\times {{10}^{9}}J\] energy must be expended to rocket the satellite out of the earth's gravitational influence.

Ans: Mass of each star, \[M=2\times {{10}^{30}}kg\]

Radius of each star, \[R={{10}^{4}}km={{10}^{7}}m\]

Distance between the stars, \[r={{10}^{9}}km={{10}^{12}}m\]

For negligible speeds, \[v=0\] 

Total energy of two stars separated at distance \[r\] is given by

\[TE=\frac{-GMM}{r}+\frac{1}{2}m{{v}^{2}}\]

\[\Rightarrow TE=\frac{GMM}{r}+0.\]……(i)

Now, consider the case when the stars are about to collide:

Velocity of the stars \[=v\]

Distance between the centres of the stars \[=2R\]

Total kinetic energy of both stars \[=\frac{1}{2}M{{v}^{2}}+\frac{1}{2}M{{v}^{2}}=M{{v}^{2}}\]

Total potential energy of both stars\[=\frac{-GMM}{2R}\]

Total energy of the two stars \[=M{{v}^{2}}-\frac{GMM}{2R}\]........(ii)

Using the law of conservation of energy, we can write: 

\[M{{v}^{2}}-\frac{GMM}{2R}=\frac{-GMM}{R}\]

\[\Rightarrow {{v}^{2}}=\frac{-GM}{r}+\frac{GM}{2r}=GM\left( -\frac{1}{r}+\frac{1}{2R} \right)\]

\[\Rightarrow {{v}^{2}}=6.67\times 1{{0}^{-10}}\times 2\times {{10}^{30}}\left[ -\frac{1}{{{10}^{12}}}+\frac{1}{2\times {{10}^{7}}} \right]\]  

\[\Rightarrow {{v}^{2}}=13.34\times {{10}^{19}}\left[ -{{10}^{-12}}+5\times {{10}^{-8}} \right]\]

\[\Rightarrow {{v}^{2}}\sim 13.34\times {{10}^{19}}\times 5\times {{10}^{-8}}\]

\[\Rightarrow {{v}^{2}}\sim 6.67\times {{10}^{12}}\]

\[\Rightarrow v=\sqrt{6.67\times {{10}^{12}}}\]

\[\Rightarrow v=2.58\times {{10}^{6}}m/s\]

The speed with which the two stars collide is \[2.58\times {{10}^{6}}m/s\].