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

A computational representation of Gravitational force can be created using VPython.

The code below show how we can find the net force, momentum and final position in Python.

while t < 235920:
      rate(5000)
      #Calculate the Gravitational Force acting on the craft due to Earth and Moon

Gravitational Interaction between the space craft, Earth and the Moon. The red arrow represents the momentum of the spacecraft at the moment

      #Craft to Earth Calculations
      r = craft.pos - Earth.pos
      rmag = sqrt(r.x**2 + r.y**2 + r.z**2)
      fgravmag = (G * mEarth * mcraft) / (rmag**2
      rhat = r / rmag
      fgrav = -1 * fgravmag * rhat
      #Craft to Moon Calculations
      rmc = craft.pos - Moon.pos
      rmcmag = sqrt(rmc.x**2 + rmc.y**2 + rmc.z**2)
      fgravmoonmag = (G * mMoon * mcraft) / (rmcmag**2)
      rmchat = rmc / rmcmag
      fgravmoon = -1 * fgravmoonmag * rmchat
      #Fnet is sum of these two forces
      fnet = fgrav + fgravmoon

From this we can calculate the change in momentum and the new positions based on the forces.

      #Update the position of the spacecraft by calculating the new momentum and the new velocity
      pcraft = pcraft + (fnet*deltat)
      vcraft = pcraft / mcraft
      craft.pos = craft.pos + (vcraft*deltat)

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 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

• Problems taken from Textbook and old Test on T-square

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 in an airplane at 40,000 feet above earth's surface. This would place the student 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}}} = \lt 0,0,0\gt m }[/math]- Position of the Sun

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

[math]\displaystyle{ \vec{\mathbf{r_{M}}} }[/math] = <[math]\displaystyle{ L,h,0\gt }[/math] m - 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}} }[/math] = <[math]\displaystyle{ L,h,0\gt }[/math]-<[math]\displaystyle{ L,0,0\gt }[/math]

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

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

Connectedness

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2. How is it connected to your major?
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History

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See also

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Further reading

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External links

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References

[1] "Newton's Law of Universal Gravitation." Newton's Law of Universal Gravitation. N.p., n.d. Web. 27 Nov. 2015.<http://www.physicsclassroom.com/Class/circles/u6l3c.cfm> [2]