Mars with a period of How (in km) above the surface the planet is (The mass of Mars is 6.42 X rbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the orbiter tracing a circle around the planet at a certain height, the sp this with the circular velocity equation to determine the height above the planet's surface. GM GM s and solving for r gives the following equation. What is the exponent for r? annot come back) kg, and the radius of Mars is
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
![**Text Explanation for Educational Website:**
**MAVEN Orbiter's Orbit Analysis**
The MAVEN orbiter travels around Mars with an orbital period of 4.5 hours. To determine how far above Mars' surface the orbiter is, we need to consider Mars' mass and radius, which are \(6.42 \times 10^{23}\) kg and \(3.40 \times 10^3\) km, respectively.
**Part 1 of 3**
The period of the orbiter's orbit allows us to calculate the velocity of the orbiter as it circles Mars. Assuming a circular orbit, the velocity \( v \) is given by the formula:
\[
v = \frac{2\pi r}{P}
\]
**Part 2 of 3**
To find the height above Mars' surface, we integrate the formula for circular velocity with the gravitational equation:
\[
v = \sqrt{\frac{GM}{r}}
\]
Equating and rearranging the formulas, we have:
\[
\frac{2\pi r}{P} = \sqrt{\frac{GM}{r}}
\]
By squaring both sides and solving for \( r \), the equation becomes:
\[
r = \frac{GM}{4\pi^2} P^2
\]
This equation helps calculate the distance \( r \) from the center of Mars, factoring in the planet's gravitational influence and the orbital period.
**Instructions:**
Complete the calculation and determine the exponent for \( r \).
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![Congratulations! You just derived a version of Kepler's Third Law for Mars!
Using the mass of Mars in kilograms and converting the 4.5 hours to seconds, calculate the distance from the center of the planet.
GM kg
4π²
]s)²
3 =
And then determine the distance (in km) from the surface.
r = rm + rs
rs
km
=
km](https://content.bartleby.com/qna-images/question/b60d532a-9aac-45a2-a466-49da1bb79676/41a8c1d7-a443-4ba7-b6d4-ef52634fd78a/c4v2d09_processed.png)
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