Chapter 7: Gravitation

Ans: The situation is represented in the given figure:

Mass of each sphere, \[M=100kg\]

Separation between the spheres, \[r=1m\]

\[X\] is the midpoint between the spheres. Gravitational force at point \[X\] will be zero. This is because the gravitational force applied by each sphere will act in opposite directions.

Gravitational potential at point \[X\]\[=\frac{-GM}{\left( \frac{r}{2} \right)}-\frac{GM}{\left( \frac{r}{2} \right)}\]

\[\Rightarrow PE=-4\frac{GM}{r}\]

\[\Rightarrow PE=\frac{4\times 6.67\times {{10}^{-11}}\times 100}{r}\]

\[\Rightarrow PE=-2.67\times {{10}^{-8}}J/kg\]

Any object placed at point \[X\] will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will vary the effective force in that direction.

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

Radius of the Earth, \[R=6400km=6.4\times {{10}^{6}}m\]

Height of a geostationary satellite from the surface of the Earth,

\[h=36000km=3.6\times {{10}^{7}}m\]

Gravitational potential energy due to Earth’s gravity at height \[h\],

\[PE=\frac{-GM}{(R+h)}\]

\[\Rightarrow PE=-\frac{6.67\times 1{{0}^{-11}}\times 6.0\times {{10}^{24}}}{3.6\times {{10}^{7}}+0.64\times {{10}^{7}}}\]

\[\Rightarrow PE=-\frac{6.67\times 6}{4.24}\times {{10}^{13-7}}\]

\[\Rightarrow PE=-9.4\times {{10}^{6}}Jk{{g}^{-1}}\]

The potential energy due to earth's gravity at the site of this satellite is \[-9.4\times {{10}^{6}}Jk{{g}^{-1}}\].

Ans: Yes.

A body gets held to the star's surface if the inward gravitational force is larger than the outward centrifugal force generated by the star's rotation.

Gravitational force, 

\[{{f}_{g}}=\frac{GMm}{{{R}^{2}}}\]

Where, M is mass of the star, \[M=2.5\times 2\times {{10}^{30}}=5\times {{10}^{30}}kg\]

m is the mass of the body

R is the radius of star, \[\text{R}=12km=1.2\times {{10}^{4}}m\]  

\[\therefore {{f}_{g}}=\frac{6.67\times {{10}^{-11}}\times 5\times {{10}^{30}}\times m}{{{(1.2\times {{10}^{4}})}^{2}}}\]

\[{{f}_{g}}=2.31\times {{10}^{11}}mN\]

Centrifugal force, \[{{f}_{c}}=mr{{\omega }^{2}}\]

\[\omega =\text{Angular speed}=2\pi v\]

\[v=\text{Angular frequency}=1.2rev\text{ }{{s}^{-1}}\]

\[{{f}_{c}}=mR{{(2\pi v)}^{2}}\]

\[\Rightarrow {{f}_{c}}=m\times (1.2\times {{10}^{4}})\times 4\times {{(3.14)}^{2}}\times {{(1.2)}^{2}}\]

\[\Rightarrow {{f}_{c}}=1.7\times {{10}^{5}}mN\]

Since \[{{f}_{g}}>{{f}_{c}}\], the body will remain held to the surface of the star.

Ans: Mass of the spaceship, \[{{m}_{s}}=1000kg\]

Mass of the Sun, \[M=2\times 1{{0}^{30}}kg\]

Mass of Mars, \[{{M}_{m}}=6.4\times {{10}^{23}}kg\]

Orbital radius of Mars, \[R=2.28\times {{10}^{8}}kg=2.28\times {{10}^{11}}m\]

Radius of Mars, \[r=3395km=3.395\times {{10}^{6}}m\]

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

Potential energy of the spaceship due to the gravitational attraction of the Sun\[=\frac{-GM{{m}_{s}}}{R}\]

Potential energy of the spaceship due to the gravitational attraction of Mars\[=\frac{-G{{M}_{m}}{{m}_{s}}}{r}\]

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.

Total energy of the spaceship \[=\frac{-GM{{m}_{s}}}{R}-\frac{G{{M}_{m}}{{m}_{s}}}{r}\]

\[TE=-G{{m}_{s}}\left( \frac{M}{R}+\frac{{{M}_{m}}}{r} \right)\] 

The negative sign indicates that the system is in a bound state.

Energy required for launching the spaceship out of the solar system = – (Total energy of the spaceship)

We have,

\[E=G{{m}_{s}}\left( \frac{M}{R}+\frac{{{m}_{m}}}{r} \right)\]

\[\Rightarrow E=6.67\times 1{{0}^{-11}}\times {{10}^{3}}\times \left( \frac{2\times {{10}^{30}}}{2.28\times {{10}^{11}}}+\frac{6.4\times {{10}^{23}}}{3.395\times {{10}^{5}}} \right)\]

\[\Rightarrow E=6.67\times {{10}^{-8}}\left( 87.72\times {{10}^{17}}+1.88\times {{10}^{17}} \right)\]

\[\Rightarrow E=6.67\times {{10}^{-8}}\times 1.88\times {{10}^{17}}\]

\[\Rightarrow E=596.97\times {{10}^{9}}\]

\[\Rightarrow E=6\times {{10}^{11}}J\]

The required energy for the spaceship to launch it out of the solar system is \[6\times {{10}^{11}}J\].

Ans: Initial velocity of the rocket, \[v=2km/s=2\times {{10}^{3}}m/s\]

Mass of Mars, \[M=6.4\times {{10}^{23}}kg\]

Radius of Mars, \[R=3395km=3.395\times {{10}^{6}}m\]

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

Mass of the rocket\[=m\]

Initial kinetic energy of the rocket \[=\frac{1}{2}m{{v}^{2}}\]

Initial potential energy of the rocket \[=\frac{-GMm}{R}\]

Total initial energy \[=\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}\]

If \[20%\] of initial kinetic energy is lost due to Martian atmospheric resistance, then it shows only \[80%\] of its kinetic energy helps in reaching a height.

Total initial energy available, \[TE=\frac{80}{100}\times \frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}\]  

\[TE=0.4m{{v}^{2}}-\frac{GMm}{R}\]

Maximum height reached by the rocket\[=h\]

At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height $h,$ \[=-\frac{GMm}{\left( R+h \right)}\]

Applying the law of energy conservation for the rocket, we can write:

\[0.4m{{v}^{2}}-\frac{GMm}{R}=\frac{-GMm}{\left( R+h \right)}\]

\[\Rightarrow 0.4{{v}^{2}}=\frac{GM}{R}-\frac{GM}{R+h}\]

\[\Rightarrow 0.4{{v}^{2}}=GM\left( \frac{1}{R}-\frac{1}{R+h} \right)\]

\[\Rightarrow 0.4{{v}^{2}}=GM\left( \frac{R+h-R}{R\left( R+h \right)} \right)\]

After solving further, we get

\[0.4{{v}^{2}}=\frac{GMh}{R\left( R+h \right)}\]

\[\Rightarrow \frac{R+h}{h}=\frac{GM}{0.4{{v}^{2}}R}\]

\[\Rightarrow \frac{R}{h}+1=\frac{GM}{0.4{{v}^{2}}R}\]

\[\Rightarrow \frac{R}{h}=\frac{GM}{0.4{{v}^{2}}R}-1\]

\[\Rightarrow h=\frac{R}{\frac{GM}{0.4{{v}^{2}}R}-1}\]

\[\Rightarrow h=\frac{0.4{{R}^{2}}{{v}^{2}}}{GM-0.4{{v}^{2}}R}\]

Putting all the values, we get,  

\[h=\frac{0.4\times {{\left( 3.395\times {{10}^{6}} \right)}^{2}}\times {{\left( 2\times {{10}^{3}} \right)}^{2}}}{6.67\times 1{{0}^{-11}}\times 6.4\times {{10}^{23}}-0.4\times {{\left( 2\times {{10}^{3}} \right)}^{2}}\times \left( 3.395\times {{10}^{6}} \right)}\]

\[\Rightarrow h=\frac{18.442\times {{10}^{18}}}{42.688\times {{10}^{12}}-53432\times {{10}^{12}}}\]

\[\Rightarrow h=\frac{18.442}{37.256}\times {{10}^{6}}\]

\[\Rightarrow h=495\times {{10}^{3}}m\]

\[\Rightarrow h=495km\]

So, the height attained by the rocket from the surface will be $495km$.