To solve this problem, we need to calculate the gravitational pull on a 200 kg satellite orbiting Earth at a distance of \( \frac{3R}{2} \), where \( R \) is the radius of Earth. The gravitational force experienced by an object in orbit can be calculated using Newton's law of universal gravitation:
\(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\)
Where:
On Earth's surface, the gravitational force per kilogram is given as 10 N. Therefore, we can find the mass of the Earth multiplied by the gravitational constant (GM) by using the gravitational force formula at the surface:
\(g = \frac{G \cdot m_1}{R^2} = 10 \, \mathrm{N/kg}\)
Thus, \(GM = 10R^2\).
For the satellite, the radius of the orbit is given as \(r = \frac{3R}{2}\). Therefore, the gravitational force on the satellite is:
\(F = \frac{G \cdot m_1 \cdot m_2}{\left(\frac{3R}{2}\right)^2} = \frac{G \cdot m_1 \cdot m_2}{\frac{9R^2}{4}} = \frac{4 \cdot G \cdot m_1 \cdot m_2}{9R^2}\)
Substitute \(GM = 10R^2\) and \(m_2 = 200 \, \mathrm{kg}\):
\(F = \frac{4 \times 10 \times 200}{9} = \frac{8000}{9} \approx 888.89 \, \mathrm{N}\)
Rounding off gives approximately \(889 \, \mathrm{N}\), which means the correct answer is 889 N.
Therefore, the gravitational pull on the satellite is 889 N.
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)