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Harrizon [31]
2 years ago
10

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.
Physics
1 answer:
ikadub [295]2 years ago
6 0

Answer:

  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

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