E>0, KE> U, object escapes potential well. E=0, KE = U, object just reaches edge of well. E =0 E< 0, KE< |U], object remains in potential well. E=-JU|, KE=0, object remains at bottom of well Total energy is at min of potential E =-|U|
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Gravitational Equilibrium: A stable orbit of an object of mass m around another object of mass M requires that the energy of motion or “kinetic” energy of the object be a certain proportion of its gravitational binding energy or “potential” energy. The kinetic energy, KE, is defined
KE=1/2mv^2
where m is the mass of the object and v is its speed. The gravitational potential energy is defined as
U=-(GMm)/(r)
where G is the gravitational constant of Newton’s so-called Universal Law of Gravitation, M and m are the masses of the interacting objects and r is the radius of the orbit. The total energy, E, of the object of mass m is the sum of its kinetic and potential energies, E = KE + U. Note the minus sign in the definition of potential energy. This can be interpreted as the amount of energy an object has to overcome in order to escape the gravitational influence of another object. (Imagine rolling a marble up the surface of the inside of a bowl - see below. The well of the bowl represents the negative gravitational binding or potential energy. If the marble is not given sufficient kinetic energy, it remains inside the bowl - i.e. bound to the gravitational potential).
Use the Universal Law of Gravitation and Newton’s Second Law of motion, along with the aforementioned definitions of kinetic and potential energy to develop the algebraic relationship between kinetic and potential energy for a circular orbit (this is also known as the Virial Theorem).
![E>0, KE> U, object
escapes potential well.
E=0, KE = U, object
just reaches edge of
well.
E =0
E< 0, KE< |U], object
remains in potential
well.
E=-JU|, KE=0, object
remains at bottom of
well Total energy is at
min of potential
E =-|U|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea3e086a-7335-48be-8aee-f374ecb6ff85%2Ffd8efb3f-f660-425d-870c-39cb3ed9dd0a%2F599fosw.png&w=3840&q=75)
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