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Wize High School Grade 12 Physics Textbook > Dynamics

Gravitational Force

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Gravitational Force


When we talk about gravity, we always talk about it as a force that comes from the Earth equal to your mass multiplied by "g". But this is actually a specific case of a more general law: Newton's Law of Universal Gravity.







Gravity is an attractive force between every pair of massive object. The full form of Newtonian gravity is




Fg=Gm1m2r2\boxed{F_g=\dfrac{Gm_1m_2}{r^2}}



- where G is the Gravitational constant G=6.677×10−11Nm2/kg2G=6.677\times10^{-11}Nm^2/kg^2
- r is the distance between the centres of mass of the two objects

  • This is an example of an Inverse Square Law: The force is inversely proportional to the square of the distance between objects


  • If we used Newton's second law, we can get a general form for the gravitational acceleration from some mass M.

a=GMr2\boxed{a=\dfrac{GM}{r^2}}


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Several notes

  1. This is a force, which is a vector. By Newton's third law, the forces on the two objects must have equal magnitudes but opposite directions. For example, the Earth is pulling down us as much as we are pulling up on it (same force)! But because of how big Earth is, that force doesn't accelerate it at all.
  2. Look at how ridiculously small G is. That's why for everyday objects, the gravitational force is negligible.

Watch Out!
It's very common for students to just use the radius of the planets/stars for r. This is wrong! Do not do that. The radius here is the distance between centres of masses (m1 and m2).


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The Special Case of Earth


Let's look at the special case, right on the surface of the Earth.



We need the radius (RE=6.371 ×106mR_E=6.371\ \times10^6m) and the mass of the Earth (5.972 ×1024kg5.972\ \times10^{24}kg ). Let's say you are mass mm , then the force on you is

Fg=GMERE2 m = (6.677×10−11 Nm2kg2) 5.972 ×1024 kg(6.371 ×106 m)2 ×m =9.81 Nkg ×m F_g=\frac{GM_E}{R_E^2}\ m\ =\ \left(6.677\times10^{-11}\ \frac{Nm^2}{kg^2}\right)\ \frac{5.972\ \times10^{24}\ kg}{\left(6.371\ \times10^6\ m\right)^2}\ \times m\ =9.81\ \frac{N}{kg}\ \times m\

Good thing the dimensions all work out here! Newton divided by kilogram is just an acceleration (m/s2). We have recovered g = 9.81m/s2g\ =\ 9.81m/s^2 .





Wize Tip
In general, you can always use this general form for gravitational force/acceleration. But if you are "close enough" to the surface, just use the constant "g". It's usually obvious when you need to use the general form, but I would say 9.81m/s2 is good up to about 10,000 m above the surface.



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Example: Losing Weight at High Altitudes!


The Earth has mass 5.927 x 1024 kg and radius 6371 km. How far above the Surface of the Earth would a 50 kg person have to go in order for his weight to be 15 % less than it is at the surface?

Solution:

m=5.927×1024kgm=5.927\times10^{24}kg

RE=6371 kmR_E=6371\text{ }km
Let's find the weight (Force of gravity) right on the surface and at some new distance from the center of earth RfR_f

FG surface=Gm1MERE2F_{G\text{ }surface}=\dfrac{Gm_1M_E}{R_E^2}

Ffar=Gm1ME(Rf)2F_{far}=\dfrac{Gm_1M_E}{(R_f)^2}
Based on the question stem information, weight at the new position should be 15 percent less than the weight on the surface of earth. So, it should be 85 percent of the weight on the surface of earth:

Ffar=0.85FsurfF_{far}=0.85F_{surf}

Gm1ME(Rf)2=Gm1MERE2(0.85)\dfrac{\cancel{Gm_1M_E}}{(R_f)^2}=\dfrac{\cancel{Gm_1M_E}}{R_E^2}(0.85)

Rf2=RE20.85R_f^2=\dfrac{R_E^2}{0.85}

Rf=6910 kmR_f=6910\text{ }km

Now we can find the height respect to the Earth surface by subtracting what we found by the radius of Earth:

Δr=6910−6371=539.32 km farther\Delta r=6910-6371=539.32\text{ }km\text{ }farther

Three identical point masses, each of 3.41 kg are located at the corners of an equilateral triangle of side length 1.5 m. What is the net force on one of the masses?

Practice: Newton's Law of Universal Gravitation

The Earth has mass 5.927 x 1024 kg and radius 6371 km. How far above the Surface of the Earth would a 50 kg person have to go in order for his weight to be 15 % less than it is at the surface?

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Acceleration Due to Gravity on Different Planets

Earth has radius RE (6371 km) and mass ME (5.97*1024 kg). The gravitational acceleration at the surface is given by g=GMR2g=G\frac{M}{R^2}(G = 6.67*10-11 Nm2/kg2).

What would the gravitational acceleration on the following planets be? Give your answer in terms of g.
a) Mercury (M = 0.055ME, R = 2440 km)
b) Jupiter (M = 317.8ME, R = 69900 km)