Respuesta :
Answer:
Explanation:
Expression for escape velocity
ve = [tex]\sqrt{\frac{2GM}{R} }[/tex]
ve² R / 2 = GM
M is mass of the planet , R is radius of the planet .
At distance r >> R , potential energy of object
= [tex]\frac{-GMm}{r}[/tex]
Since the object is at rest at that point , kinetic energy will be zero .
Total mechanical energy = [tex]\frac{-GMm}{r}[/tex] + 0 = [tex]\frac{-GMm}{r}[/tex]
Putting the value of GM = ve² R / 2
Total mechanical energy = ve² Rm / 2 r
This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy . So at surface of the earth , total mechanical energy
= ve² Rm / 2 r
The total mechanical energy of the particle of mass m at the surface of the planet of mass M when the distance is R (planet's radius) is [tex] E = \frac{GmM}{R} + \frac{1}{2}mv_{e}^{2} [/tex].
When the particle is at rest at a distance "r" from the planet's center, it only has gravitational potential energy. As the particle falls to the planet's surface, some initial gravitational potential energy converts to kinetic energy.
The total mechanical energy of the particle at the planet's surface is equal to the sum of the gravitational potential energy and kinetic energy, so:
[tex] E = P + K [/tex]
[tex] E = maR + \frac{1}{2}mv_{e}^{2} [/tex] (1)
Where:
m: is the mass of the object
a: is the acceleration due to gravity on the surface of the planet
R: is the radius of the planet
[tex]v_{e}[/tex]: is the escape velocity
The acceleration due to gravity is given by:
[tex] a = \frac{GM}{R^{2}} [/tex] (2)
By entering equation (2) into (1), we have:
[tex] E =mR\frac{GM}{R^{2}} + \frac{1}{2}mv_{e}^{2} [/tex]
[tex] E = \frac{GmM}{R} + \frac{1}{2}mv_{e}^{2} [/tex]
Therefore, the total mechanical energy of the particle is [tex] E = \frac{GmM}{R} + \frac{1}{2}mv_{e}^{2} [/tex]
Learn more here:
- https://brainly.com/question/9322440?referrer=searchResults
- https://brainly.com/question/14127859?referrer=searchResults
I hope it helps you!
