Newton’s Law of Universal Gravitation
Solved Problems
Example 1.
A small cosmic body starts to fall to Earth from rest under the action of gravitational force. The initial distance to the centre of the Earth is equal to \(L.\) Determine the velocity and time of the drop.
Solution.
The motion of the body occurs along a straight line towards the centre of the Earth. Given that the body weight is much smaller than the mass of the Earth, the differential equation describing its motion can be written as
\[\frac{{{d^2}r}}{{d{t^2}}} = - G\frac{{{M_\text{E}}}}{{{r^2}}},\]
where \({M_\text{E}}\) is the mass of the Earth.
This is a nonlinear equation of type \(y^{\prime\prime} = f\left( y \right),\) which allows reduction of order. Given that
\[\frac{{{d^2}r}}{{d{t^2}}} = \frac{{dv}}{{dt}} = \frac{{dv}}{{dr}}\frac{{dr}}{{dt}} = v\frac{{dv}}{{dr}},\]
the equation takes the form:
\[v\frac{{dv}}{{dr}} = - G\frac{{{M_\text{E}}}}{{{r^2}}}.\]
Integrate it by separating the variables with the initial condition \(v\left( {r = L} \right) = 0:\)
\[vdv = - G{M_\text{E}}\frac{{dr}}{{{r^2}}},\;\; \Rightarrow \int {vdv} = - G{M_\text{E}}\int {\frac{{dr}}{{{r^2}}}} ,\;\; \Rightarrow \frac{{{v^2}}}{2} = \frac{{G{M_\text{E}}}}{r} + {C_1},\;\; \Rightarrow v = \sqrt {\frac{{2G{M_\text{E}}}}{r} + {C_1}} .\]
Given the initial condition, we have:
\[0 = \sqrt {\frac{{2G{M_\text{E}}}}{L} + {C_1}} ,\;\; \Rightarrow {C_1} = - \frac{{2G{M_\text{E}}}}{L},\;\; \Rightarrow v = \sqrt {2G{M_\text{E}}\left( {\frac{1}{r} - \frac{1}{L}} \right)} .\]
In the limiting case as \(L \to \infty,\) the formula for the velocity is simplified to
\[v = \sqrt {\frac{{2G{M_\text{E}}}}{r}} .\]
This expression can be rewritten through the gravitational acceleration \(g = {\frac{{G{M_\text{E}}}}{{R_\text{E}^2}}},\) where \({R_\text{E}}\) is the radius of the Earth. Then
\[v = \sqrt {\frac{{2G{M_\text{E}}}}{r}} = \sqrt {\frac{{2gR_\text{E}^2}}{r}} .\]
Hence, assuming that the body moves from infinity, the velocity when it hits the ground is
\[v\left( {r = {R_\text{E}}} \right) = \sqrt {\frac{{2gR_\text{E}^2}}{{{R_1}}}} = \sqrt {2g{R_\text{E}}} ,\]
that is will be equal to the escape velocity \(v \approx 10,2\,\frac{\text{km}} {\text{s}}.\)
For a finite value of \(L\), the terminal velocity of the body is less than the escape velocity:
\[v\left( {r = {R_\text{E}}} \right) = \sqrt {2G{M_\text{E}}\left( {\frac{1}{{{R_\text{E}}}} - \frac{1}{L}} \right)} = \sqrt {2gR_\text{E}^2\left( {\frac{1}{{{R_\text{E}}}} - \frac{1}{L}} \right)} = \sqrt {2g{R_\text{E}}\left( {1 - \frac{{{R_\text{E}}}}{L}} \right)} = \sqrt {2g{R_\text{E}}} \sqrt {1 - \frac{{{R_\text{E}}}}{L}} .\]
We now determine the time of falling of the body, assuming that the initial distance to the centre of the Earth is equal to \(L.\) As \({\frac{{dr}}{{dt}}} = - v,\) we obtain the following differential equation describing the motion of the body along the radial axis:
\[\frac{{dr}}{{dt}} = - {R_\text{E}}\sqrt {2g} \sqrt {\frac{1}{r} - \frac{1}{L}} ,\;\; \Rightarrow \frac{{dr}}{{\sqrt {\frac{1}{r} - \frac{1}{L}} }} = - {R_\text{E}}\sqrt {2g} dt,\]
where the distance \(r\) varies from \(L\) to \({R_\text{E}}.\)
To integrate this equation, we change the variable:
\[\frac{1}{r} - \frac{1}{L} = {z^2},\;\; \Rightarrow \frac{1}{r} = {z^2} + \frac{1}{L},\;\; \Rightarrow - \frac{1}{{{r^2}}}dr = 2zdz,\;\; \Rightarrow \frac{1}{{{r^2}}}dr = - 2zdz,\;\; \Rightarrow {\left( {{z^2} + \frac{1}{L}} \right)^2}dr = - 2zdz,\;\; \Rightarrow dr = - \frac{{2zdz}}{{{{\left( {{z^2} + \frac{1}{L}} \right)}^2}}}.\]
Then the equation can be rewritten as
\[- \frac{{2zdz}}{{{{\left( {{z^2} + \frac{1}{L}} \right)}^2}z}} = - {R_\text{E}}\sqrt {2g} dt,\;\; \Rightarrow 2\int {\frac{{dz}}{{{{\left( {{z^2} + \frac{1}{L}} \right)}^2}}}} = {R_\text{E}}\sqrt {2g} t + C.\]
It is known that the resulting integral is given by
\[\int {\frac{{dx}}{{{{\left( {{x^2} + {a^2}} \right)}^2}}} = \frac{x}{{2{a^2}\left( {{x^2} + {a^2}} \right)}} + \frac{1}{{2{a^3}}}\arctan \frac{x}{a}.}\]
Therefore, in our case we have
\[\int {\frac{{dz}}{{{{\left( {{z^2} + \frac{1}{L}} \right)}^2}}}} = \frac{z}{{\frac{2}{L}\left( {{z^2} + \frac{1}{L}} \right)}} + \frac{{{L^{\frac{3}{2}}}}}{2}\arctan \left( {z\sqrt L } \right).\]
Substituting the integral, we can write the equation as
\[2\left[ {\frac{z}{{\frac{2}{L}\left( {{z^2} + \frac{1}{L}} \right)}} + \frac{{{L^{\frac{3}{2}}}}}{2}\arctan \left( {z\sqrt L } \right)} \right] = {R_\text{E}}\sqrt {2g} t + C,\;\; \Rightarrow \frac{{zL}}{{{z^2} + \frac{1}{L}}} + {L^{\frac{3}{2}}}\arctan \left( {z\sqrt L } \right) = {R_\text{E}}\sqrt {2g} t + C.\]
Let us turn back from the variable \(z\) to the variable \(r:\)
\[\frac{{L\sqrt {\frac{1}{r} - \frac{1}{L}} }}{{\frac{1}{r}}} + {L^{\frac{3}{2}}}\arctan \left( {\sqrt L \sqrt {\frac{1}{r} - \frac{1}{L}} } \right) = {R_\text{E}}\sqrt {2g} t + C,\;\; \Rightarrow r\sqrt L \sqrt {\frac{L}{r} - 1} + {L^{\frac{3}{2}}}\arctan \sqrt {\frac{L}{r} - 1} = {R_\text{E}}\sqrt {2g} t + C.\]
Given the initial condition \(r\left( {t = 0} \right) = L,\) we find that the constant \(C\) is zero. Since at the time when the body hits the ground \(r\left( {t = T} \right) = {R_\text{E}},\) we obtain the following expression for the fall time \(T:\)
\[{R_\text{E}}\sqrt L \sqrt {\frac{L}{{{R_\text{E}}}} - 1} + {L^{\frac{3}{2}}}\arctan \sqrt {\frac{L}{{{R_\text{E}}}} - 1} = {R_\text{E}}\sqrt {2g} T,\;\; \Rightarrow T = \frac {1}{{R_\text{E}}\sqrt {2g}} {\left[{ {R_\text{E}}\sqrt L \sqrt{\frac{L}{{{R_\text{E}}}} - 1} + {L^{\frac{3}{2}}}\arctan \sqrt {\frac{L}{{{R_\text{E}}}} - 1} }\right]}.\]
After some algebra, the final exact formula for the fall time can be written as
\[T = \sqrt {\frac{L}{{2g}}} \left[ {\sqrt {\frac{L}{{{R_\text{E}}}} - 1} + \frac{L}{{{R_\text{E}}}}\arctan \sqrt {\frac{L}{{{R_\text{E}}}} - 1} } \right].\]
For large values of the ratio \({\frac{L}{{{R_\text{E}}}}}\) (in this case the \(\arctan\) function tends to \(\frac{\pi }{2}\)), we obtain the simple expression:
\[T \approx\; \sqrt {\frac{L}{{2g}}} \left[ {\sqrt {\frac{L}{{{R_\text{E}}}} - 1} + \frac{L}{{{R_\text{E}}}}\frac{\pi }{2}} \right] \approx \frac{\pi }{2}{\sqrt {\frac{L}{{2g}}} \frac{L}{{{R_\text{E}}}}.}\]
For example, the fall time from a distance of \(100,000\;\text{km}\) is estimated to be
\[T \approx\; \frac{\pi }{2}\sqrt {\frac{L}{{2g}}} \frac{L}{{{R_\text{E}}}} = \frac{\pi }{2} \sqrt {\frac{{{{10}^8}}}{{2 \cdot 10}}} \cdot \frac{{{{10}^8}}}{{6,4 \cdot {{10}^6}}} \approx 54,900\;\text{sec} = 15,2\;\text{hours}.\]
