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The substitution you should make is that $mg = \frac{GM_{\rm E}m}{R^2}$ where $g$ is the value of the gravitational field strength at a distance $R$ from the centre of the Earth. Note that the value of $g$ is not constant.
m 1 = the mass of object that's doing the escaping (a spacecraft, explosion debris, meteorite impact debris, etc.) m 2 = the mass of the astronomical object it's escaping from (the Earth, sun, Milky Way, etc.) v 0 = the initial speed of the escaping object: v = the speed of the escaping object at some later time: r 0 =
In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. [1][2] The term potential energy was introduced by the 19th-century Scottish engineer and physicist William Rankine, [3][4][5] although it has links to the ancient Greek ph...
10 Οκτ 2023 · Calculate the unknown variable in the equation for gravitational potential energy, where potential energy is equal to mass multiplied by gravity and height; PE = mgh. Calculate GPE for different gravity of different enviornments - Earth, the Moon, Jupiter, or specify your own. Free online physics calculators, mechanics, energy, calculators.
PE grav = m *• g • h In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.
The gravitational potential energy of an object is the 'stored energy' that the object has by being at that height. This is equivalent to its mass times the force of gravity, g (a defined constant of 9.8 m/s 2) times the height of the object. Potential energy = mass x gravity x height. Egrav = PE = mgh.
16 Αυγ 2021 · The change in gravitational potential energy \(\Delta PE_g\), is \(\Delta PE_g = mgh\), with \(h\) being the increase in height and \(g\) the acceleration due to gravity. The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system.
