A mass M is split into two parts, m and M % m, which are then separated by a certain distance. What ratio m/M maximizes the magnitude of the gravitational force between the parts?
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Question
Three dimensions. Three point particles are fixed in place in an xyz coordinate system. Particle A, at the origin, has mass \(m_A\). Particle B, at xyz coordinates (2.00d, 1.00d, 2.00d), has mass \(2.00 m_{A}\), and particle C, at coordinates (-1.00d, 2.00d, -3.00d), has mass \(3.00 m_{A}\). A fourth particle D, with mass \(4.00 m_{A}\), is to be placed near the other particles. In terms of distance d, at what
(a) x,
(b) y, and
(c) z coordinate should D be placed so that the net gravitational force on A from B, C, and D is zero?
Solution
Step 1 of 5
The figure below shows the arrangement of the three particles in the xyz coordinate system.
Here, A, B, and C are the three particles in the xyz coordinate system.
Particle A is located at the origin of the xyz coordinate system. Using the Pythagorean Theorem, the position of each particle from the origin or point A is as follows.
\(r = \sqrt {{x^2} + {y^2} + {z^2}}\)
Here, x, y, and z are the components of each particle.
The distance of the particle B from the particle A is,
\({r_B} = \sqrt {{{\left( {2d} \right)}^2} + {{\left( {1d} \right)}^2} + {{\left( {2d} \right)}^2}} \)
\({r_B} = 3d\)
The distance of the particle C from the particle A is,
\({r_C} = \sqrt {{{\left( { - d} \right)}^2} + {{\left( {2d} \right)}^2} + {{\left( { - 3d} \right)}^2}} \)
\({r_C} = \sqrt {14} d\)
The x component of the force exerted on particle A by the particle B is,
\({\left( {{F_{AB}}} \right)_x} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{x_B}\)
Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A,\( {m_B}\) is the mass of the particle B, and \({x_B}\) is the x component of the position of the particle B.
Substitute 2.00\({m_A}\) for \({m_B}\), 3d for \({r_B}\), and 2.00d for \({x_B}\) in the above expression, and we get,
\({\left( {{F_{AB}}} \right)_x} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {2.00d} \right)\)
\({\left( {{F_{AB}}} \right)_x} = \frac{{4Gm_A^2}}{{27{d^2}}}\)
The y component of the force exerted on particle A by the particle B is,
\({\left( {{F_{AB}}} \right)_y} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{y_B}\)
Here, \( {y_B}\) is the y component of the position of the particle B.
Substitute 2.00\({m_A}\) for \({m_B}\), 3d for \({r_B}\), and 1.00d for \({y_B}\) in the above expression, and we get,
\({\left( {{F_{AB}}} \right)_y} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {1.00d} \right)\)
\({\left( {{F_{AB}}} \right)_y} = \frac{{2Gm_A^2}}{{27{d^2}}}\)
The z component of the force exerted on particle A by the particle B is,
\({\left( {{F_{AB}}} \right)_z} = \frac{{G{m_A}{m_B}}}{{r_B^3}}{z_B}\)
Here, \({z_B}\) is the z component of the position of the particle B.
Substitute 2.00\({m_A}\) for \({m_B}\), 3d for \({r_B}\), and 2.00d for \({z_B}\) in the above expression, and we get,
\({\left( {{F_{AB}}} \right)_z} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {3d} \right)}^3}}}\left( {2.00d} \right)\)
\({\left( {{F_{AB}}} \right)_z} = \frac{{4Gm_A^2}}{{27{d^2}}}\)
The x component of the force exerted on particle A by the particle C is,
\({\left( {{F_{AC}}} \right)_x} = \frac{{G{m_A}{m_C}}}{{r_C^3}}{x_C} \)
Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A, \({m_C}\) is the mass of the particle C, and \({x_C}\) is the x component of the position of the particle C.
Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d for \({r_C}\), and -1.00d for \({x_C}\) in the above expression, and we get,
\({\left( {{F_{AC}}} \right)_x} = \frac{{G{m_A}3.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( { - 1.00d} \right)\)
\({\left( {{F_{AC}}} \right)_x} = \frac{{ - 3\sqrt {14} Gm_A^2}}{{196{d^2}}}\)
The y component of the force exerted on particle A by the particle C is,
\({\left( {{F_{AC}}} \right)_y} = \frac{{G{m_A}{m_C}}}{{r_C^3}}{y_C}\)
Here, \({y_C}\) is the y component of the position of the particle C.
Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d for \({r_C}\), and -1.00d for \({y_C}\) in the above expression, and we get,
\({\left( {{F_{AC}}} \right)_y} = \frac{{G{m_A}3.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( {2.00d} \right)\)
\({\left( {{F_{AC}}} \right)_y} = \frac{{3\sqrt {14} Gm_A^2}}{{98{d^2}}}\)
The z component of the force exerted on particle A by the particle C is,
\({\left( {{F_{AC}}} \right)_z} = \frac{{G{m_A}{m_B}}}{{r_C^3}}{z_C}\)
Here, \({z_C}\) is the z component of the position of the particle B.
Substitute 3.00\({m_A}\) for \({m_C}\), \(\sqrt {14}\)d for \({r_C}\), and -1.00d for \({z_C}\) in the above expression, and we get,
\({\left( {{F_{AC}}} \right)_z} = \frac{{G{m_A}2.00{m_A}}}{{{{\left( {\sqrt {14} d} \right)}^3}}}\left( { - 3.00d} \right)\)
\({\left( {{F_{AC}}} \right)_z} = \frac{{ - 9\sqrt {14} Gm_A^2}}{{196{d^2}}}\)
The x component of the force exerted on particle A by the particle D is,
\(\left( {{F_{AD}}} \right) = \frac{{G{m_A}{m_D}}}{{r_D^2}} \)
Here, G is the universal gravitational constant, \({m_A}\) is the mass of the particle A,\( {m_D}\) is the mass of the particle D, and \({r_D}\) is the distance between points A and D.
Substitute 4.00\({m_A}\) for \({m_D}\) in the above expression, and we get,
\(\left( {{F_{AD}}} \right) = \frac{{G{m_A}4.00{m_A}}}{{r_D^2}}\)
\(\left( {{F_{AD}}} \right) = \frac{{4Gm_A^2}}{{r_D^2}}\)
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