Gravity Calculation on a New Planet
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.
Comparing Planet Gravity to Earth
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:
- Mass of the planet: \(M_p = \frac{M_E}{2}\)
- Radius of the planet: \(R_p = \frac{R_E}{2}\)
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:
\(g^1 = 2g\)
Conclusion on Planet Gravity
The acceleration due to gravity on the surface of this new planet is twice the acceleration due to gravity on Earth.
Important Questions from Universal law of gravitation
Which of the following laws says that "Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them?"
The force of attraction between two objects of masses 'M' and 'm' which lie at a distance 'd' from each other is directly proportional to the-
The force of attraction (F) between two particles having masses m 1and m 2is given by _______. (If r is the distance between them and G is a universal constant)
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.
Imagine a light planet is revolving around a star in a circular orbit of radius R with the period of revolution T . If the gravitational force of attraction between the two is proportional to R(-5/2) then
