Gravitational Force

Main Idea

A Mathematical Model

What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Units

In Si Unit, Gravitational Force F is measured in Newtons (N), the two masses, m₁ and m₂ are measures in kilograms (Kg), the distance is measured in meters (m), and the gravitational constant G is measured in N m²/ kg⁻² and has a value of 6.674×10⁻¹¹ N m²/ kg⁻². The Gravitational Constant G have different values for different units. The value of constant G appeared in Newton's law of universal gravitation, but it was not measured until seventy two years after Newton's death by Henry Cavendish with his Cavendish experiment in 1798. The value of gravitational constant was also the first test of Newton's law between two masses in Laboratory.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Determine the force of gravitational attraction between the earth (m = [math]\displaystyle{ 6 x 10^{24} kg }[/math]) and a 70 kg physics student if the student is standing at sea level, a distance of [math]\displaystyle{ 6.38 x 10^6 m }[/math] from earth's center.

The solution of the problem involves substituting known values of G ([math]\displaystyle{ 6.673 x 10^{-11} }[/math] N m2/kg2), m₁ ([math]\displaystyle{ 6 x 10^{24} }[/math] kg), m₂ (70 kg) and d ([math]\displaystyle{ 6.38 x 10^6 m }[/math]) into the universal gravitation equation and solving for Fgrav. The solution is as follows:

[math]\displaystyle{ |\mathbf{F_{g}}| = G \frac{m_E m_M}{r^2} }[/math]

[math]\displaystyle{ |\mathbf{F_{g}}| = 6.7x10^{-11} \frac{6e24 * 70}{6.38 x 10^6} }[/math]

[math]\displaystyle{ |\mathbf{F_{g}}| =686 N }[/math]

Middling

The mass of the Earth is [math]\displaystyle{ 6 x 10^ {24} }[/math] kg, and the mass of the Moon is [math]\displaystyle{ 7 x 10^ {22} }[/math] kg. At a particular instance the moon is at location [math]\displaystyle{ \lt 2.8 x 10^8,0,-2.8 x 10^8\gt }[/math] m, in a coordinate system whose origin is at the center of the earth. (a) What is [math]\displaystyle{ \vec{\mathbf{r}} }[/math], the relative position vector from the Earth to the Moon? (b) What is [math]\displaystyle{ |\vec{\mathbf{r}}| }[/math]? (c) What is the unit vector [math]\displaystyle{ \vec{\mathbf{r}} }[/math]? (d) What is the gravitation force exerted by the Earth on the Moon? Your answer should be in vector.

The solution of the problem involves substituting known values of G ([math]\displaystyle{ 6.673 x 10^{-11} }[/math]N m²/kg²), m₁ [math]\displaystyle{ 6 x 10^ {24} }[/math] kg, m₂ [math]\displaystyle{ 7 x 10^ {22} }[/math] kg and d [math]\displaystyle{ \lt 2.8 x 10^8,0,-2.8 x 10^8\gt }[/math] m m into the universal gravitation equation and solving for F_grav. The solution is as follows:

Gravitational Force Between Moon and Earth

(a) The position vector of Moon relative to Earth is,

[math]\displaystyle{ \vec{\mathbf{r}} }[/math] = [math]\displaystyle{ \lt 2.8 x 10^8,0,-2.8 x 10^8\gt - \lt 0,0,0\gt }[/math]

[math]\displaystyle{ \vec{\mathbf{r}} }[/math] = [math]\displaystyle{ \lt 2.8 x 10^8,0,-2.8 x 10^8\gt }[/math] m

(b) The magnitude of position vector of Moon relative to Earth is,

[math]\displaystyle{ |\vec{\mathbf{r}}| }[/math] = [math]\displaystyle{ \sqrt{(2.8 x 10^8)^2+0^2+(-2.8 x 10^8)^2} }[/math]

[math]\displaystyle{ |\vec{\mathbf{r}}| }[/math] = [math]\displaystyle{ 4.0 x 10^8 }[/math] m

(c) The unit vector of Moon relative to Earth is,

[math]\displaystyle{ {\mathbf{r}} = \frac {\vec{\mathbf{r}}}{|\vec{\mathbf{r}}|} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \frac{\lt 2.8 x 10^8,0,-2.8 x 10^8\gt }{4.0 x 10^8} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \lt 0.7,0,-0.7\gt }[/math]

(d) The expression for the gravitational force on the Moon by the Earth is,

[math]\displaystyle{ |\vec{\mathbf{F}}_{g}| = G \frac{m_E m_M}{|\vec{\mathbf{r^2}}|} }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g}| = 6.7x10^{-11} \frac{6 x 10^ {24} * 7 x 10^ {22}}{(4.0 x 10^8)^2 } }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g}| = 1.76x10^{20}N }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g} }[/math] = [math]\displaystyle{ -|\mathbf{F_{grav}}|*{\mathbf{r}} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g} }[/math] = [math]\displaystyle{ -1.76x10^{20}*\lt 0.7,0,-0.7\gt N }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g} }[/math] = [math]\displaystyle{ \lt -1.232x10^{20},0,1.232x10^{20}\gt N }[/math]

Difficult

In the following problems you will be asked to calculate the net gravitational force acting on the Moon.To do so, please use the following variables:

Mass

m_S - Mass of the Sun

m_E - Mass of the Earth

m_M - Mass of the Moon

Initial Positions

[math]\displaystyle{ \vec{\mathbf{r_{S}}} }[/math] = [math]\displaystyle{ \lt 0,0,0\gt m }[/math] - Position of the Sun

[math]\displaystyle{ \vec{\mathbf{r_{E}}} }[/math] = [math]\displaystyle{ \lt L,0,0\gt m }[/math] - Position of the Earth

[math]\displaystyle{ \vec{\mathbf{r_{M}}} }[/math] = [math]\displaystyle{ \lt L,h,0\gt m }[/math] - Position of the Moon

(a) Calculate the gravitational force on the Moon due to the Earth (b) Calculate the gravitational force on the Moon due to the Sun (c) Calculate the net gravitational force on the Moon

(a) The gravitational force on the Moon due to the Earth is,

[math]\displaystyle{ \vec{\mathbf{r}} = \vec{\mathbf{r_M}}-\vec{\mathbf{r_E}} }[/math]

[math]\displaystyle{ \vec{\mathbf{r}} = \lt L,h,0\gt -\lt L,0,0\gt }[/math]

[math]\displaystyle{ \vec{\mathbf{r}} = \lt 0,h,0\gt m }[/math]

[math]\displaystyle{ |\vec{\mathbf{r}}| = \sqrt{(0^2+h^2+0^2)} }[/math]

[math]\displaystyle{ |\vec{\mathbf{r}}| = h }[/math] m

[math]\displaystyle{ {\mathbf{r}} = \frac {\vec{\mathbf{r}}}{|\vec{\mathbf{r}}|} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \frac{\lt 0,h,0\gt }{h} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \lt 0,1,0\gt }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g_1}| = G \frac{m_E m_M}{|\vec{\mathbf{r^2}|}} }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g_1}| = G \frac{m_E m_M}{h^2} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g_1} = -|\vec{\mathbf{F}}_{g_1}|*{\mathbf{r}} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g_1} = \lt 0,-G \frac{m_E m_M}{h^2},0\gt N }[/math]

(b) The gravitational force on the Moon due to the Sun is,

[math]\displaystyle{ \vec{\mathbf{r}} = \vec{\mathbf{r_M}}-\vec{\mathbf{r_S}} }[/math]

[math]\displaystyle{ \vec{\mathbf{r}} = \lt L,h,0\gt -\lt 0,0,0\gt }[/math]

[math]\displaystyle{ \vec{\mathbf{r}} = \lt L,h,0\gt m }[/math]

[math]\displaystyle{ |\vec{\mathbf{r}}| = \sqrt{(L^2+h^2+0^2)} }[/math]

[math]\displaystyle{ |\vec{\mathbf{r}}| = \sqrt{(L^2+h^2)} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \frac {\vec{\mathbf{r}}}{|\vec{\mathbf{r}}|} }[/math]

[math]\displaystyle{ {\mathbf{r}} = \frac{\lt L,h,0\gt }{\sqrt{(L^2+h^2)}} }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g_2}| = G \frac{m_S m_M}{|\vec{\mathbf{r^2}|}} }[/math]

[math]\displaystyle{ |\vec{\mathbf{F}}_{g_2}| = G \frac{m_S m_M}{L^2+h^2} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g_2} = -|\vec{\mathbf{F}}_{g_2}|*{\mathbf{r}} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{g_2} = \lt -G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0\gt N }[/math]

(c) The net gravitational force on the Moon is,

[math]\displaystyle{ \vec{\mathbf{F}}_{net} = \vec{\mathbf{F}}_{g_1}+\vec{\mathbf{F}}_{g_2} }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{net} = \lt 0,-G \frac{m_E m_M}{h^2},0\gt +\lt -G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0\gt }[/math]

[math]\displaystyle{ \vec{\mathbf{F}}_{net} = \lt -G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_E m_M}{h^2}-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0\gt N }[/math]

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