A satellite circles the earth in an orbit whose radius is 2.77 times the earth's radius. The earth's mass is 5.98 x 1024 kg, and its radius is 6.38 x 106 m. What is the period of the satellite?
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.
![**Question:**
A satellite circles the earth in an orbit whose radius is 2.77 times the earth's radius. The earth's mass is \(5.98 \times 10^{24} \text{ kg}\), and its radius is \(6.38 \times 10^{6} \text{ m}\). What is the period of the satellite?
**Answer:**
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**Resources:**
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**Explanation:**
To solve the problem of finding the period of a satellite orbiting the Earth, one can use Kepler’s Third Law for circular orbits. The law states that the square of the orbital period \( T \) is proportional to the cube of the semi-major axis of its orbit \( r \):
\[ T^2 \propto r^3 \]
\[ T^2 = \frac{4 \pi^2 r^3}{G M} \]
Where:
- \( T \) is the orbital period
- \( r \) is the radius of the orbit, which in this case is 2.77 times the Earth’s radius \((2.77 \times 6.38 \times 10^6 \text{ m})\)
- \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}) \)
- \( M \) is the mass of the Earth \( (5.98 \times 10^{24} \text{ kg}) \)
**Calculation Steps:**
1. Determine the orbital radius \(r\).
2. Plug values into the proportional formula to compute the period \(T\).
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