IDNLearn.com offers a comprehensive solution for all your question and answer needs. Our platform offers comprehensive and accurate responses to help you make informed decisions on any topic.
Sagot :
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
