Respuesta :
Answer:
A. 6419 km
B. 9.822 m/s²
Explanation:
Parameters given:
Mass of earth, M = 5.974 * 10^24 kg
Radius of earth, R = 6371 km = 6.371 * 10^6 m
Gravitational constant, G = 6.674 * 10^(-11) Nm²/kg²
Orbital period, T = 4 hours = 14400 secs
A. Height of the satellite above the earth's surface can be found by subtracting the earth's radius, R, from the radius of the orbit of the satellite, Ro.
The radius of the orbit, Ro, is given as:
Ro = ³√(GMT²/4π²)
Ro = ³√[(6.674 * 10^(-11) * 5.974 * 10^24 * 14400²)/4π²]
Ro = ³√[(8.268 * 10²¹)/3.948]
Ro = ³√(2.084 * 10²¹)
Ro = 1.279 * 10^7 m = 12790 km
Height = Ro - R
Height = 12790 - 6371
Height = 6419 km
B. Orbital acceleration:
The orbital acceleration is equal to the acceleration due to gravity. This can be proven with the formula
g = GM/R²
g = (6.674 * 10^(-11) * 5.974 * 10^24)/(6.371 * 10^6)²
g = 9.822 m/s²
The height from the surface of the earth at which the satellite is rotating is 6427 km.
The acceleration of the satellite is [tex]9.297 \;\rm m/s^2[/tex]
Given that, a satellite moving around the earth in circular orbit completes one revolution in 4.00 h.
Mass of Earth is 5.974 × 1024 kg, radius of Earth is 6371 km and Gravitational constant G is = 6.674 × 10−11 N·m2/kg2.1)
Let us consider ‘R’ is the distance between the earth’s center and the satellite, 'r' is the radius of the earth and 'h' is the height from the surface of the earth at which the satellite is rotating. Then, the distance 'R' can be written as,
[tex]R =r+h[/tex]
Now the distance R can also be calculated as,
[tex]R=\sqrt[3]{\dfrac {GMT^2}{4\pi^2}}[/tex]
Substituting the values of G, M and T, the distance R is,
[tex]R=\sqrt[3]{\dfrac {6.674\times 10^{-11}\times 5.974\times10^{24}\times(4\times60\times60)^2}{4\times3.14^2}}[/tex]
[tex]R=\sqrt[3]{\dfrac {8.268\times10^{22}}{39.438}}[/tex]
[tex]R = \sqrt[3]{2.096\times 10^{21}}[/tex]
[tex]R= 12797655.8207[/tex]
[tex]R=12797.65\;\rm km[/tex]
So the height from the surface of the earth at which the satellite is rotating can be calculated as,
[tex]h=R-r[/tex]
[tex]h=12797.65-6371[/tex]
[tex]h=6426.65 \;\rm km[/tex]
The height from the surface of the earth at which the satellite is rotating is 6427 km.
The satellite's acceleration is calculated given below.
[tex]a = \dfrac {GM}{r^2}[/tex]
As the satellite moving around the earth, hence its acceleration is equivalent to the gravitational acceleration of the earth.
[tex]a = \dfrac{6.674\times10^{-11}\times5.974\times10^{24}}{({6371\times10^3})^2}[/tex]
[tex]a= 9.297 \;\rm m/s^2[/tex]
The acceleration of the satellite is [tex]9.297 \;\rm m/s^2[/tex].
For more details, follow the link given below.
https://brainly.com/question/2069814.
