A Mars orbiter is orbiting Mars at 35,000 m above the Martian surface. What is its period in seconds? The mass of Mars is 6.42*10^23kg. The radius of Mars is 3.40*10^6m. Gravitational constant G=6.67*10^-11 N*(m^2)/(kg^2) Hint: you need the distance from the satellite to the center of Mars, not the surface of Mars for your formula.
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.
![### Orbital Mechanics Problem: Calculating the Orbit period of a Mars Orbiter
A Mars orbiter is orbiting Mars at 35,000 m above the Martian surface. What is its period in seconds?
#### Given Data:
- Mass of Mars (M): \( 6.42 \times 10^{23} \) kg
- Radius of Mars (R): \( 3.40 \times 10^{6} \) m
- Gravitational constant (G): \( 6.67 \times 10^{-11} \) N*(m²)/kg²
**Hint:** The distance from the satellite to the center of Mars, not just the surface, is required for your formula.
##### Solution Steps:
1. **Calculate Distance from Center of Mars to the Orbiter:**
The orbiter is at an altitude of 35,000 m above the Martian surface.
\[
\text{Altitude (h)} = 35,000\text{ m}
\]
Total distance, \(d\), from the center of Mars to the orbiter:
\[
d = \text{Radius of Mars (R)} + \text{Altitude (h)}
\]
\[
d = 3.40 \times 10^{6}\text{ m} + 35,000\text{ m} = 3.435 \times 10^{6}\text{ m}
\]
2. **Use Kepler’s Third Law to Find Orbital Period (T):**
Kepler’s Third Law relates the orbital period \(T\) to the distance \(d\) and the mass \(M\) of the planet:
\[
T^2 = \frac{4\pi^2 d^3}{GM}
\]
Solving for \(T\):
\[
T = \sqrt{\frac{4\pi^2 d^3}{GM}}
\]
3. **Substitute Known Values:**
\[
d = 3.435 \times 10^{6}\text{ m}
\]
\[
G = 6.67 \times 10^{-11} \text{ N*(m²)/kg²}
\]
\[
M = 6.42 \times 10^{23](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2932a05-99f3-41c1-989e-7dbca32c3377%2F0673213c-dd4a-4001-8858-c4f67015a9da%2Fxpaf15_processed.png&w=3840&q=75)
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