Posts Tagged ‘Earth’s orbital velocity’

Earth’s High Speed Race Around the Sun

Tuesday, July 28th, 2015

      Last time we began our discussion on velocity by focusing on one of its aspects, distance, and we calculated Earth’s orbital distance around the sun to be 5,816,023,200 miles.   Today we’ll focus on velocity’s other aspect, time.   Together, these aspects will allow us to solve for Earth’s orbital velocity, aka speed.

      When early astronomers monitored Earth’s journeys, they found it took exactly one year for it to complete its orbit around the sun.   They combined this fact with Earth’s orbital travel distance of 5,816,023,200, or 9.36 × 1011 meters — meters being the unit of measurement most often used in scientific computations — and solved for Earth’s orbital velocity as follows,

v = 9.36 × 1011 meters ÷ 1 year = 9.36 × 1011 meters per year

      The scientific notation of 9.36 × 1011 equates to 936,000,000,000 meters, a large and unruly number to work with.   We can simplify things further by breaking this number down into units of meters per second, which will then allow us to arrive at Earth’s velocity in terms of miles per hour, something most Earthlings can relate to.

      One meter per second is equal to 2.237 miles per hour, and there are 31,536,000 seconds in one year, so breaking Earth’s orbital velocity down into meters per second we arrive at,

v = 9.36 × 1011 meters ÷ 31,536,000 seconds = 29,680 meters per second

v = (29,680 meters per second) × (2.237 mph/m/sec) = 66,394 miles per hour

      Yes, it’s true, Earth whips around the sun at warp speed.  Our fastest man made rockets only achieve speeds of about 25,000 miles per hour.   Earth beats them two-to-one!

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      Now that we know Earth’s orbital velocity, we have everything we need to calculate the gravitational force exerted on Earth by the sun.   We’ll do that next time.


Earth’s Orbital Velocity

Sunday, July 19th, 2015

      Last time we introduced Newton’s equation to calculate the sun’s gravitational force acting upon Earth, and today we’ll begin solving for the last remaining unsolved variable within that equation, v, Earth’s orbital velocity.

      Here again is Newton’s equation,

Fg = [m × v2] ÷ r

      For a refresher on how we solved for m, Earth’s mass, and r, the distance between Earth and the sun, follow these links to past blogs in this series, What is Earth’s Mass and Calculating the Distance to the Sun.

      Velocity, or speed, as it’s most commonly referred to, is based on both time and distance.   To bear this out we’ll use an object and situation familiar to all of us, traveling in a car.   The car’s velocity is a factor of both the distance traveled and the time it takes to get there.  A car traveling at a velocity of 30 miles per hour will cover a distance of 30 miles in one hour’s time.   This relationship is borne out by the formula,

vCar = distance traveled ÷  travel time

vCar = 30 miles ÷ 1 hour = 30 miles per hour

      Similarly, v is the distance Earth travels during its orbital journey around the sun within a specified period of time.   It had been observed since ancient times that it takes Earth one year to complete one orbit, so all that remained to be done was calculate the distance Earth traveled during that time.   Vital to calculations was the fact that Earth’s orbit is a circle, which allows geometry to be employed and calculations to be thereby simplified.   Refer to Figure l.

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Figure 1

      From geometry we know that the circumference of a circle, C, is calculated by,

C = 2× π × r

where π is a constant, the well known mathematical term pi, which is equal to 3.1416, and r is the radius of Earth’s circular orbit, determined, courtesy of the work of Johannes Kepler and Edmund Halley, to be approximately 93,000,000 miles.   Stated in metric units, the unit of measurement most often employed in science, that comes to 149,000,000,000 meters.

      Inserting these numerical values for π and r into the circumference formula, scientists calculated the distance Earth travels in one orbit around the sun to be,

C = 2 ×  π × 149,000,000,000 meters = 9.36 x 1011 meters

      Next time we’ll introduce the time element into our equations and solve for v, and from there we’ll go on and finally solve for Fg, the sun’s gravitational force acting upon Earth.