Question

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2 .What is the orbital period of a spacecraft in a low orbit near the surface of Mars? Assume the radius of the satellite's orbit is about the same as the radius of Mars itself, rMars = 3.37

Answer 1

The orbital period of the spacecraft is about 5.9 × 10³ s ≈ 99 minutes

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Further explanation

Let's recall Centripetal Force formula as follows:

$\boxed{F = m \frac{v^2}{R}}$

where:

F = Centripetal Force ( Newton )

m = mass of object ( kg )

v = speed of object ( m/s )

R = radius ( m )

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Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\boxed {F = G \frac{m_1 ~ m_2}{R^2}}$

where:

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 !

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Given:

free-fall acceleration = g = 3.8 m/s²

radius of Mars = R = 3.37 × 10⁶ m

Asked:

orbital period = T = ?

Solution:

$\Sigma F = ma$

$G \frac{ M m} { R^2 } = m \omega^2 R$

$G \frac{ M } { R^2 } = \omega^2 R$

$g = \omega^2 R$

$\omega^2 = g \div R$

$\omega = \sqrt { g \div R }$

$2 \pi \div T = \sqrt { g \div R }$

$T = 2 \pi \div \sqrt { g \div R$

$T = 2 \pi \sqrt{ R \div g}$

$T = 2 \pi \sqrt{ 3.37 \times 10^6 \div 3.8 }$

$\boxed{T \approx 5.9 \times 10^3 \texttt{ s}}$

$\boxed{T \approx 99 \texttt{ minutes}}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Answer 2

Answer: 5952 s  

The equation to calculate the period (in seconds) of a body orbiting a massive body (in this specific case a spacecraft with mass $m$ orbiting Mars) is:  

$T=2\pi\sqrt{\frac{{r}^{3}}{GM}}$    (1)  

Where:  

$r$ is the radius of the orbit measured from the center of Mars to the satellite  

$G$ is the gravity constant with a value of $6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$  

$M$ is the mass of Mars, which is known to be $6.39({10}^{23})kg$  

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 $R_{MARS}$:  

$r\approx R_{MARS}=3.37({10}^{6})m$  

Having this stated, let’s begin with the solution:  

$T=2\pi\sqrt{\frac{{(3.37({10}^{6})m)}^{3}}{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})(6.39({10}^{23})kg)}}$  

Solving this equation being careful with the units, and taking into account $1N=kg\frac{m}{{s}^{2}}$ we have:  

$T=2\pi\sqrt{8.974({10}^{5}){s}^{2}}$  

Finally:  

$T=5952.1367s\approx 5952s$ >>>>>This is the period of the orbit of the spacecraft around Mars