“Concern for man himself and his fate must always

form the chief interest of all technical endeavors…Never forget this in the

midst of your diagrams and equations.”

Albert Einstein

Gravity Math

The History

• We have learned the history of gravity, the missteps and false starts, the names of the movers and shakers in the field, and the progression of the ideas

The Math

• The math for gravity is a mix of the equations for circular motion and the physics for acceleration

What could we want to know?

• Let us make a list of the possible bits of information we could want to know about an object orbiting around another, or about the attraction between two objects

The List

• Constants

• Acceleration

• Force

• Period

• Velocity

• Mass of either object

The Constant

• There is one constant involved, the gravitational constant

2

211

kg

mN1067.6G

Acceleration

• We know that a change in direction requires an acceleration

• We saw this for circular motion

Acceleration

• r is the distance from the center of the mass

• M is the mass of the attracting object

2r

GMa

Note…

• The mass of the falling object is not a factor

• Only the mass of the planet, Sun, or other large, attracting body and distance of the item from the center counts

Example 1

• At sea-level the radius of the Earth is 6.38 × 106 m. If the mass of the Earth is 5.98 × 1024 kg, what is gravitational acceleration at sea-level? What is it atop Mt. Everest, 8848 m above sea-level?

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• r1 = 6.38 × 106 m

• m = 5.98 × 1024 kg

• r2 = ((6.38 × 106)+ 8848) m

Answer 1

• a = Gm/r2

• a1 = 9.799 m/s2 ≈ 9.80 m/s2

• a2 = 9.772 m/s2 ≈ 9.77 m/s2

Example 2

• Jupiter orbits the sun at a radius of 7.78 ×1011 m. The Sun has a mass of 1.99 ×1030 kg. What is the acceleration holding Jupiter in its orbit?

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• r = 7.78 ×1011 m

• m = 1.99 ×1030 kg

Answer 2

• a = Gm/r2

• a = 2.1929 × 10-4 m/s2 ≈ 2.19 × 10-4 m/s2

Force

• We already know the basic equation

maF

Force

• If we know the mass of the object, we need only find the acceleration to calculate the force

Force

2r

GMmF

Force

• G is the gravitational constant

• M is the larger of the two masses

• m is the smaller of the two

• r is the radius between them

Force

• This was Newton’s big achievement

• He was able to find out what Kepler’s constant was

Example 3

• Pluto has a mass of 1.25 × 1022 kg and an orbital radius of 5.87 × 1012 m. What is the force, in Newtons, between Pluto and the Sun?

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• r = 5.87 × 1012 m

• m = 1.25 ×1022 kg

• M = 1.99 ×1030 kg

Answer 3

• F = GMm/r2

• F = 4.82 × 1016 N

Period

GM

rT

3

2

Period

• Remember, this is the time it takes to make one full revolution

• The time is measured in seconds

Example 4

• What is the orbital period of Pluto, given the information from the pervious problems?

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• r = 5.87 × 1012 m

• M = 1.99 ×1030 kg

Answer 4

• T = 2π(r3/(GM))1/2

• T = 7.76 × 109 s

Converted to years

• T = 246 years

Example 5

• The Earth has a period of 365 days. If it orbits at a radius of 1.50×1011 m, then what is the mass of the Sun.

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• r = 1.50 × 1011 m

• T = 365 days = 3.15 × 107 s

Answer 5

• T = 2π(r3/(GM))1/2

• M = (4π2r3)/(GT2)

• M = 2.01 × 1030 kg

Velocity

r

GMv

Velocity

• The velocity of the orbiting object depends only on the mass of the larger body, not the mass of the orbiting object

Example 6

• A satellite orbits the Earth at a distance of 340000 m. The mass of the Earth is 5.97 × 1024 kg. What velocity must it have to maintain that orbit?

What Do We Know

• G = 6.67 × 10-11 Nm2/kg2

• M = 5.97 × 1024 kg

• r = 340000 m

Answer 6

• v = ((GM)/r)1/2

• v = 34222 m/s ≈ 34200 m/s