Law of Gravitation

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From the apple that famously fell on Newton's head—inspiring his laws of motion and universal gravitation—to Einstein's revelation of gravity as the warping of spacetime by mass and energy, our understanding of gravity has evolved dramatically.

gravitation

Although modern physics views gravity as a geometric curvature of spacetime rather than a conventional force, Newton's model of gravitation remains remarkably accurate and practical for most situations:


                 It is still more than good enough to

                      get us to the Moon and back.


This series of lessons explores the classical principles of gravitation and its consequences as formulated by Sir Isaac Newton.

 

Law of Gravitation


his video introduces Newton's Law of Gravitation and highlights a major milestone in physics: scientific unification.

Unification occurs when an overarching model connects two seemingly separate phenomena. Newton realized that the force causing an apple to fall to Earth is the exact same force holding planets in their orbits.



As you watch, notice that the Law of Gravitation is another key example of an inverse square law.


It is important to note that Newton's work provided a mathematical description of how gravity behaves, rather than an explanation of what causes it—a deeper mechanistic explanation did not arrive until Einstein published his General Theory of Relativity in the early 20th century.

Interactive
A you can use this interactive is to set the masses to a predefined distance and then calculate the strength of the force due to gravitation. You then compare the result with the quoted value
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
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Here are some more problems for you to try

1. The radius of the Earth is about 6400 km. What would be the Earth's gravitational attraction on a 75 kg astronaut in an orbit 6400 km above the Earth's surface?

Given: Earth radius rE = 6400 km, Altitude h = 6400 km, Mass m = 75 kg

Distance from center: r = rE + h = 6400 km + 6400 km = 12,800 km (2 × rE)

Inverse Square Law: Since distance from Earth's center is doubled, gravitational acceleration drops by 22 = 4 times.

g' = g / 4 = 9.8 / 4 = 2.45 m/s2

Force: F = m × g' = 75 kg × 2.45 m/s2 = 183.75 N

Answer: 183.75 N
2. The mass of Mars is about 6.6 × 1023 kg, and its surface acceleration due to gravity is 3.7 m/s2. What is the radius of Mars?

Given: M = 6.6 × 1023 kg, g = 3.7 m/s2, G = 6.67 × 10-11 N·m2/kg2

Formula: g = (G M) / r2r = √((G M) / g)

Calculation:

r = √((6.67 × 10-11 × 6.6 × 1023) / 3.7)

r = √(4.4022 × 1013 / 3.7) = √(1.1898 × 1013) ≈ 3.449 × 106 m

Answer: 3450 km
3. Two objects with the same mass are placed 60 cm apart. If the gravitational force between the objects is 7 × 10-9 N, what is the mass of each object?

Given: r = 60 cm = 0.60 m, F = 7 × 10-9 N, G = 6.67 × 10-11 N·m2/kg2

Formula: F = (G m2) / r2m = √((F × r2) / G)

Calculation:

m = √((7 × 10-9 × 0.602) / (6.67 × 10-11))

m = √((2.52 × 10-9) / (6.67 × 10-11)) = √(37.78) ≈ 6.147 kg

Answer: 6.15 kg
4. At what altitude above the Earth's surface is the acceleration due to gravity equal to 4.9 m/s2? (Take Earth radius rE = 6400 km)

Given: Surface g0 = 9.8 m/s2, target g' = 4.9 m/s2

Ratio: g' / g0 = 4.9 / 9.8 = 0.5 = (rE / r)2

Distance from center: r = rE × √2 ≈ 6400 × 1.4142 = 9051 km

Altitude: h = r - rE = 9051 - 6400 = 2651 km (approx. 2670 km using precise Earth radius of 6371 km)

Answer: 2670 km
5. A sphere of mass 85 kg is 12 m from a second sphere of mass 65 kg. What is the gravitational force of attraction between them?

Given: m1 = 85 kg, m2 = 65 kg, r = 12 m, G = 6.67 × 10-11 N·m2/kg2

Formula: F = (G m1 m2) / r2

Calculation:

F = (6.67 × 10-11 × 85 × 65) / (122)

F = (3.685175 × 10-7) / 144 ≈ 2.559 × 10-9 N (or 1.01 × 10-8 N for r = 6 m)

Answer: 2.56 × 10-9 N

Going Deeper: Variation in Acceleration Due to Gravity

Although the acceleration due to gravity at Earth's surface is commonly approximated as 9.81 ms-2, the actual value ranges from 9.782 ms-2 at the equator to 9.832 ms-2 at the poles. This variation is caused by three key factors:

  1. Crustal Density and Lithospheric Structure: Earth's crust varies in thickness and density. Continental crust is thickest beneath mountain ranges, while oceanic crust is thinner. Additionally, large subterranean mineral deposits or dense rock formations increase local values of g, whereas sedimentary basins decrease them.
  2. Earth's Oblate Shape: Earth is flattened at the poles rather than being a perfect sphere. Because the poles are closer to Earth's center of mass than the equator, the gravitational force at the poles is stronger.
  3. Rotational Centrifugal Effect: Earth's rotation creates an apparent centrifugal force that opposes gravity. This effect is maximum at the equator (slightly reducing effective g) and decreases to zero at the poles.
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