The acceleration due to gravity (\(g\)) on the surface of a celestial body is determined by its mass (\(M\)) and radius (\(R\)) using the formula:
\(g = \frac{GM}{R^2}\)
where \(G\) is the universal gravitational constant.
Let \(g\) represent the acceleration due to gravity on Earth, with mass \(M_E\) and radius \(R_E\). The formula is:
\(g = \frac{GM_E}{R_E^2}\)
For the new planet, let its acceleration due to gravity be \(g^1\), its mass be \(M_p\), and its radius be \(R_p\).
We are given the following relationships:
Using the gravity formula for the new planet:
\(g^1 = \frac{GM_p}{R_p^2}\)
Substitute the given values for \(M_p\) and \(R_p\) into the equation:
\(g^1 = \frac{G \left( \frac{M_E}{2} \right)}{\left( \frac{R_E}{2} \right)^2}\)
Simplify the expression step-by-step:
\(g^1 = \frac{G \cdot M_E / 2}{R_E^2 / 4}\)
\(g^1 = \frac{G M_E}{2} \times \frac{4}{R_E^2}\)
\(g^1 = 2 \times \frac{GM_E}{R_E^2}\)
Recognize that \(\frac{GM_E}{R_E^2}\) is equal to Earth's gravity, \(g\). Substitute \(g\) back into the equation:
\(g^1 = 2g\)
The acceleration due to gravity on the surface of this new planet is twice the acceleration due to gravity on Earth.
Which of the following is NOT correct ?
Three point masses each of mass m are placed at the three corners of an equilateral triangle of side x. Find the resultant force acting on any one particle at the corner.
Which of the following is NOT correct ?