Question

An earth satellite remains in orbit at a distance of 13594 km from the center of the earth. what is its period? the universal gravitational constant is 6.67 × 10−11 n · m2 /kg2 and the mass of the earth is 5.98 × 1024 kg.

Answer 1

To calculate the period of an earth satellite at a given distance from the center of the earth, we can use Kepler's third law.

Step-by-step explanation:

To calculate the period of an earth satellite, we can use Kepler's third law. The formula is T² = (4π²r³)/(GM). Here, T represents the period, r represents the distance of the satellite from the center of the earth, G is the gravitational constant, and M is the mass of the earth.

In this case, the satellite's distance from the center of the earth is given as 13594 km. We can convert it to meters by multiplying it by 1000. The mass of the earth is given as 5.98 x 10²⁴ kg. Plugging these values into the formula, we can calculate the period of the satellite.

Answer 2

Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of satellite - radius of the earth

r = 13,594 Km - 6,380 Km = 7,214 Km convert to meter "m"

r = 7,214,000 m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: What is the period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe

T² = 39.48(7,214,000 m)³/(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.48 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 37,092,731.83 s²

T = 6,090.38 seconds or 1.7 Hr