12. A body of weight 72 N moves from the surface of Earth at a height half the radius of Earth. What is the gravitational force on it due to Earth?
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Beyond the Answer
To calculate the gravitational force on a body at a height equal to half the radius of the Earth, you can use the formula for gravitational force, which states that the force decreases with the square of the distance from the center of the Earth. The gravitational force \( F \) at a distance \( r \) from the center can be defined as: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] At the Earth's surface, the radius \( R \) is approximately 6,371 km. At a height of \( \frac{R}{2} \), the distance from the center of the Earth becomes \( \frac{3R}{2} \). The gravitational force on the body at this height will be: \[ F' = \frac{72 N}{\left(\frac{3}{2}\right)^2} = \frac{72 N}{\frac{9}{4}} = 72 N \times \frac{4}{9} = 32 N \] Thus, the gravitational force on the body at that height is 32 N. --- Did you know that Isaac Newton's law of universal gravitation laid the foundation for understanding gravity? In the late 17th century, he formulated the idea that every mass attracts every other mass in the universe, setting the stage for our understanding of everything from falling apples to the orbits of planets! In practice, engineers and scientists use gravitational calculations every day, especially in fields like aerospace. They must determine the gravitational forces on rockets, satellites, and even space stations as they operate at various altitudes. Knowing how gravity changes with altitude ensures that these vehicles can navigate the vast expanse of space safely and efficiently!
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