Modeled after the two satellites problem (slide 12 in lecture13, universal gravity), imagine now if the period of the satellite is exact 88.59 hours, the earth mass is 5.98 x 1024 kg, and the radius of the earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles? use g=6.67x 10 -11 nm2/kg2. your answer could be a large number on the order of tens of thousands, just put in the raw number you get, for example, 12345.67.

  R = 6.3456 10⁴  mile

Explanation:

For this exercise we will use Newton's second law where force is gravitational force

      F = m a

The satellite is in a circular orbit therefore the acceleration is centripetal

      a = v² / r

Where the distance is taken from the center of the Earth

     G m M / r² = m v² / r

     G M / r = v²

The speed module is constant, let's use the uniform motion relationships, with the length of the circle is

     d = 2π  r

     v = d / t

The time for a full turn is called period (T)

Let's replace

     G M / r = (2π r / T)²

     r³ = G M T²²2 / 4π²

     r = ∛ (G M T² / 4π²)

We have the magnitudes in several types of units

      T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s

      Re = 6.37 10⁶ m

Let's calculate

     r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)

     r = ∛ (1.027487 10²⁴)

     r = 1.0847 10⁸ m

This is the distance from the center of the Earth, the distance you want the surface is

     R = r - Re

     R = 108.47 10⁶ - 6.37 10⁶

     R = 102.1 10⁶ m

Let's reduce to miles

      R = 102.1 10⁶ m (1 mile / 1609 m)

      R = 6.3456 10⁴  mile

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr