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Wize University Physics Textbook (Master) > Newton's Laws of Motion: Forces and 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?
Extra Practice
g on different planets
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.
Practice: Newton’s Law of Universal Gravitation
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?
Kepler's third law example
Practice: Satellite (Orbital Period)
Some bright students are planning to send a satellite to orbit around the earth at an altitude of 4000.0 km above the surface. What will be the orbital period of this satellite? (answer in seconds)
Practice: Satellite Orbiting Earth
A 600 kg satellite and a 450 kg satellite are in orbit around the Earth. The 600kg satellite is at a distance of 2000 km from the Earth’s surface while the 450 kg satellite is at a distance of 15 000 km.
a) What is period of the satellites?
Kepler's third law example
Some bright students are planning to send a satellite to orbit around the earth at altitude of 4000 km above the surface. What will be the orbital period of this satellite?
Because of Earth's rotation about its axis, a point at the poles has no centripetal acceleration. (Assume Earth is a uniform sphere and take g=9.800 ms2. g=9.800\ \frac{m}{s^2}.\ )
A 5000kg satellite is in orbit around the Earth and has an orbital radius of 35 900 km.
a) What is the magnitude of the force that the satellite exerts on the Earth? (mass of Earth = 5.98 x 1024 kg)
b) By what proportion does the force change if the satellite's mass is cut by 75% and the orbit height cuts in half?
Three identical point masses, each of 3.41 kg are located at the corners of an equilateral triangle of side length 1.5 m.
a) Consider a point halfway along one of the sides. What would be the direction of the net force at this point if we put another mass there? (Hint: Always start physics problems by drawing a picture!)
b) What is the net force on one of the three masses? (Remember: force is a vector, so your answer should have a magnitude and a direction.)
Simple Harmonic Motion: Pendulum
You have landed on an unknown planet. You have a simple pendulum with a length of 25.0 cm25.0\ cm. You measure the frequency of oscillation of the pendulum to be 0.5 Hz0.5\ Hz.
a) What is the gravitational acceleration on this planet? (in m/s^2)
b) What would be the period of the oscillation if you used this pendulum on Earth? (in seconds)
Practice: Measuring gravity
You set up a simple experiment to measure gravitational acceleration on Earth. You drop a ball from 14 feet above the ground (a second-floor window). You use a stopwatch to find that it falls from rest in 1.013 seconds. Calculate Earth's gravitational acceleration based on these results to the appropriate number of significant figures, and comment on your result with one suggestion for improving the experiment. (one foot=0.305 meters)
Practice: Mars Rock Drop
A martian drops a rock from a height of 8.71 meters on Mars, where gravitational acceleration near the planet's surface is 3.711 m/s2.
g 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)
A 600 kg satellite and a 450 kg satellite are in orbit around the Earth. The 600kg satellite is at a distance of 2000 km from the Earth’s surface while the 450 kg satellite is at a distance of 15 000 km.
a) What is period of the satellites?
b) What is their speed?
Kepler's third law example
Some bright students are planning to send a satellite to orbit around the earth at altitude of 4000.0 km above the surface. What will be the orbital period of this satellite? (answer in seconds)
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?
Getting Earth out of orbit
The Sun is about to explode. We need to get the Earth out of the solar system.
Radius and mass of the Sun is 6.95 x 108m and 1.99 x 1030kg. Radius and mass of the Earth is 6.37 x 106m and 5.97 x 1024kg.
Escape from Mars
You are in a spaceship that's moving at 4.20 x 103m/s at an altitude of 254km above the surface of Mars. The mass of Mars is 6.46 x 1023kg and radius is 3.39 x 106m. Will you be able to get into space or will you fall back to the planet?
Satellite into space
How much energy do we need to send a 9000.0 kg satellite from the Earth to the ISS, which is at 4.00 x 105m above the Earth's surface? You are given the mass of the Earth (5.98 x 1024kg), the radius of the Earth (6.37 x 106m). Answer in Joules.
Moon's orbit
Given the mass of the Earth (5.98 x 1024kg), the radius of the Earth (6.37 x 106m), the mass of the moon (7.35 x 1022kg) and the radius of the moon (1.74 x 106m).
a) Calculate the magnitude of the acceleration of the moon due to the Earth's gravity. The distance to the moon is 3.84 x 108 m.
b) Calculate the centripetal acceleration if the moon is in a circualr orbit. The period of the moon's orbit is 2.73 days.
The acceleration due to gravity depends on the...
A particle with mass m1 = 20 kg, is at the origin, the second particle with mass m2= 30 kg is at −2ȷ^-2\hat\jmath and the 3rd mass with m3= 5 kg is at 3ı^+4ȷ^3\hat\imath+4\hat\jmath .
What is the gravitational force between m2 and m3? What is the total gravitational force on m3?
Practice: Percent Error
An experiment finds that the acceleration due to gravity is 10.2m/s2. What is the percentage error in this experiment? Use g=9.8m/s2 as the accepted value for gravity.
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?
A satellite circles the earth at an altitude of approximately 20,000 km, where the gravitational acceleration is 5.8% of its surface value. Find the orbital period of the satellite.
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?