The equation to calculate the period (in seconds) of a body orbiting a massive body (in this specific case a spacecraft with mass [tex]m[/tex] orbiting Mars) is:
[tex]T=2\pi\sqrt{\frac{{r}^{3}}{GM}}[/tex] (1)
Where:
[tex]r[/tex] is the radius of the orbit measured from the center of Mars to the satellite
[tex]G[/tex] is the gravity constant with a value of [tex]6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}[/tex]
[tex]M[/tex] is the mass of Mars, which is known to be [tex]6.39({10}^{23})kg[/tex]
Now, if we are told that the spacecraft is in a low orbit near the surface of Mars, we can assume the radius of the orbit is approximately equal to the radius of Mars [tex]R_{MARS}[/tex]:
[tex]r\approx R_{MARS}=3.4({10}^{6})m[/tex]
Having this stated, let’s begin with the solution:
[tex]T=2\pi\sqrt{\frac{{(3.4({10}^{6})m)}^{3}}{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})(6.39({10}^{23})kg) }}[/tex]
Solving this equation being careful with the units, and taking into account [tex]1N=kg\frac{m}{{s}^{2}}[/tex] we have:
[tex]T=2\pi\sqrt{9.216({10}^{5}){s}^{2}}[/tex]
Finally:
[tex]T=6031,9s[/tex] >>>>>This is the period of the orbit of the spacecraft around Mars
The orbital period of a spacecraft in a low orbit near the surface of Mars is about 5.9 × 10³ s ≈ 99 minutes
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Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
free-fall acceleration = g = 3.8 m/s²
radius of Mars = R = 3.4 × 10⁶ m
Asked:
orbital period = T = ?
Solution:
[tex]\Sigma F = ma[/tex]
[tex]G \frac{ M m} { R^2 } = m \omega^2 R[/tex]
[tex]G \frac{ M } { R^2 } = \omega^2 R[/tex]
[tex]g = \omega^2 R[/tex]
[tex]\omega^2 = g \div R[/tex]
[tex]\omega = \sqrt { g \div R }[/tex]
[tex]2 \pi \div T = \sqrt { g \div R }[/tex]
[tex]T = 2 \pi \div \sqrt { g \div R [/tex]
[tex]T = 2 \pi \sqrt{ R \div g}[/tex]
[tex]T = 2 \pi \sqrt{ 3.4 \times 10^6 \div 3.8 }[/tex]
[tex]\boxed{T \approx 5.9 \times 10^3 \texttt{ s}}[/tex]
[tex]\boxed{T \approx 99 \texttt{ minutes}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Gravitational Fields
