Respuesta :
Answer:
a)[tex]r_1=1.36R[/tex]
b)[tex]r_2=2.083R[/tex]
Explanation:
Given:
a) when the initial velocity of the projectile is 0.520 times the escape velocity from the earth.
Let r be the radial distance from the earth's surface Let M be the mass of the Earth and R be the radius of the Earth
Now using conservation of Energy at earths surface and at distance r we have
[tex]\dfrac{-GMm}{R}+\dfrac{m(0.52V_e)^2}{2}=\dfrac{-GMm}{r_1}\\\dfrac{-GMm}{R}+\dfrac{m\times 0.52^2\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_1}\\r_2=1.36\ R[/tex]
b) when the Initial kinetic Energy of the projectile is 0.52 times the Kinetic Energy required to escape the Earth
Conservation of Energy we have
[tex]\dfrac{-GMm}{R}+0.52\times KE_{escape}=\dfrac{-GMm}{r_2}\\\dfrac{-GMm}{R}+0.52\times\dfrac{m\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_2}\\r_2=2.083\ R[/tex]