Class 11 Physics - Gravitation (Important Q&A)
Gravitation Marks · 3M or 4M
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📗 3 Mark Q&A
5+ QuestionSolution:
Acceleration due to gravity at height \(h\) above the surface is \( g_1 = g \left( \dfrac{1}{1 + \dfrac{h}{R}} \right)^2 \).
Since \( R \gg h \), we use approximation \( (1+x)^{-2} \approx 1 - 2x \) for small \(x\).
So, \( g_1 \approx g \left( 1 - \dfrac{2h}{R} \right) \).
Acceleration due to gravity at depth \(h\) below the surface is \( g_2 = g \left( 1 - \dfrac{h}{R} \right) \).
\(\therefore \dfrac{g_2}{g_1} = \dfrac{ g \left( 1 - \dfrac{h}{R} \right) }{ g \left( 1 - \dfrac{2h}{R} \right) } \).
\( \Rightarrow \dfrac{g_2}{g_1} = \dfrac{ 1 - \dfrac{h}{R} }{ 1 - \dfrac{2h}{R} } \).
Using approximation \( \dfrac{1 - x}{1 - 2x} \approx 1 + x \) for small \(x\), we get
\( \dfrac{g_2}{g_1} \approx 1 + \dfrac{h}{R} \).
Hence proved, \( \dfrac{g_2}{g_1} = 1 + \dfrac{h}{R} \) when \( R \gg h \).
Solution:
Acceleration due to gravity at height \(h\) is \( g_h = g \left( \dfrac{R}{R+h} \right)^2 \).
Acceleration due to gravity at depth \(h\) is \( g_d = g \left( 1 - \dfrac{h}{R} \right) \).
Given \( g_h = g_d \).
So, \( g \left( \dfrac{R}{R+h} \right)^2 = g \left( 1 - \dfrac{h}{R} \right) \)
\( \Rightarrow \left( \dfrac{R}{R+h} \right)^2 = 1 - \dfrac{h}{R} \)
Let \( x = \dfrac{h}{R} \).
Then equation becomes \( \left( \dfrac{1}{1+x} \right)^2 = 1 - x \)
So, \( \dfrac{1}{(1+x)^2} = 1 - x \)
\( \Rightarrow 1 = (1 - x)(1+x)^2 \)
\( \Rightarrow (1 - x)(1 + 2x + x^2)=1 \)
\(\Rightarrow 1 + 2x + x^2 - x - 2x^2 - x^3=1 \)
\( \Rightarrow 1 + x - x^2 - x^3 =1 \)
\( \Rightarrow x - x^2 - x^3 = 0\)
\( \Rightarrow x(1 - x - x^2) = 0 \)
As \(x\neq 0\), we solve \( 1 - x - x^2 = 0 \)
\( x^2 + x - 1 = 0 \)
Using Sridar Acharyya's formula, we get
\( x = \dfrac{-1 \pm \sqrt{1 + 4}}{2} \)
\( \Rightarrow x = \dfrac{-1 \pm \sqrt{5}}{2} \)
Since \(x\) must be positive, \( x = \dfrac{\sqrt{5}-1}{2} \)
But \( x = \dfrac{h}{R} \)
Therefore, \( h = \left( \dfrac{\sqrt{5}-1}{2} \right) R \).
Solution:
Given that gravitational force varies inversely as the \(n\)th power of distance, so \( F \propto \dfrac{1}{r^n} \).
Let \( F = \dfrac{k}{r^n} \), where \(k\) is a constant.
For a planet of mass \(m\) moving in a circular orbit of radius \(r\), the centripetal force required is \( \dfrac{mv^2}{r} \).
Equating gravitational force to centripetal force,
\( \dfrac{mv^2}{r} = \dfrac{k}{r^n} \).
\( \Rightarrow mv^2 = \dfrac{k}{r^{n-1}} \).
\( \Rightarrow v^2 = \dfrac{k}{m r^{\,n-1}} \).
\( \Rightarrow v^2 = \dfrac{k \, r^{\,-(n-1)}}{m} \).
Hence, \( v \propto r^{-\frac{n-1}{2}} \).
Time period of revolution is \( T = \dfrac{2\pi r}{v} \).
\( \therefore T \propto \frac{r}{v} \).
Therefore, \( T \propto \dfrac{r}{r^{-\frac{n-1}{2}}} \).
\( \Rightarrow T \propto r^{1 + \frac{n-1}{2}} \).
\( \therefore T \propto r^{\frac{n+1}{2}} \).
Hence proved that \( T \propto r^{\frac{n+1}{2}} \).
Solution:
Gravitational force on the satellite is \( F = \dfrac{GMm}{r^2} \).
This provides the necessary centripetal force \( \dfrac{mv^2}{r} \).
So, \( \dfrac{mv^2}{r} = \dfrac{GMm}{r^2} \).
\( \Rightarrow \dfrac{v^2}{r} = \dfrac{GM}{r^2} \).
Hence, \( v^2 = \dfrac{GM}{r} \).
We have, time period of revolution is \( T = \dfrac{2\pi r}{v} \).
On squaring both sides, we get
\( T^2 = \dfrac{4\pi^2 r^2}{v^2} \).
On substituting \( v^2 = \dfrac{GM}{r} \), we get
\( \Rightarrow T^2 = \dfrac{4\pi^2 r^2}{\dfrac{GM}{r}} \).
\( \Rightarrow T^2 = \dfrac{4\pi^2 r^3}{GM} \).
Since \(4\pi^2\) and \(GM\) are constants,
\( \therefore T^2 \propto r^3 \).
Hence proved.
Or,
Define escape velocity. Derive the expression for it.Solution:
1st Part:
Escape Velocity It is the minimum velocity with which a body must be projected from the surface of the earth so that it just escapes the earth’s gravitational field without further propulsion.
2nd Part:
Let mass of earth be \(M\), mass of body be \(m\), and radius of earth be \(R\).
Gravitational force at a distance \(r\) from the centre of earth is \( F = \dfrac{GMm}{r^2} \).
Work done against gravity in moving the body from \(R\) to infinity is
\( W = \int_R^{\infty} \dfrac{GMm}{r^2} \, dr \).
\( \Rightarrow W = GMm \int_R^{\infty} r^{-2} \, dr \).
\( \Rightarrow W = GMm \left[ -\dfrac{1}{r} \right]_R^{\infty} \).
\( \Rightarrow W = GMm \left( 0 - \left( -\dfrac{1}{R} \right) \right) \).
\( \therefore W = \dfrac{GMm}{R} \).
This work must be supplied by the initial kinetic energy.
So, \( \dfrac{1}{2} m v_e^2 = \dfrac{GMm}{R} \).
\( \Rightarrow \dfrac{1}{2} v_e^2 = \dfrac{GM}{R} \).
\( \Rightarrow v_e^2 = \dfrac{2GM}{R} \).
\( \therefore v_e = \sqrt{\dfrac{2GM}{R}} \).
Since \(m\) cancels out, escape velocity is independent of the mass of the body.
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