Required information A spy satellite is in circular orbit around Earth. It makes one revolution in 3.30 h. Mass of Earth is 5.974 x 1024 kg, radius of Earth is 6371 km and Gravitational constant G is 6.674 x 10-11 N.m2/kg². low high above Earth's surface is the satellite? km
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.
![**Problem Statement:**
The orbital speed of Earth about the Sun is \(3.00 \times 10^4\) m/s, and its distance from the Sun is \(1.50 \times 10^{11}\) m. The mass of Earth is approximately \(6.00 \times 10^{24}\) kg, and that of the Sun is \(2.00 \times 10^{30}\) kg. What is the magnitude of the force exerted by the Sun on Earth?
\[ \_\_\_\_\_\_ \times 10^{22} \, \text{N} \]
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To solve this problem, you can use Newton's law of universal gravitation, which states that the force \(F\) between two masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by:
\[ F = G \frac{m_1 \cdot m_2}{r^2} \]
where \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\).
Given:
- \( m_1 = 6.00 \times 10^{24} \, \text{kg} \) (mass of Earth)
- \( m_2 = 2.00 \times 10^{30} \, \text{kg} \) (mass of the Sun)
- \( r = 1.50 \times 10^{11} \, \text{m} \)
Calculate \( F \) using the formula above.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86d5b315-251b-4afc-ade8-02046d912aca%2Ff27f9463-f3f4-4904-a924-a4e83b8ffd54%2Fq7d82n_processed.jpeg&w=3840&q=75)
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