Question

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.

Answer 1

R = 6.3456 10⁴ mile

Step-by-step 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