• anonymous
how to show that a body projected from the earth with a speed of square root of 2gR will never return?????. Radius of the Earth is R.AND State two assumptions for this result.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • katieb
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
Consider that you want to project the body from the surface of the Earth to infinity, say from h=0 to infinity. You need to know the required potential energy and with that calculate the kinetic energy (velocity) required. The gravitational force on the surface is:\[F_E=G \frac{ M_E ·m}{ R_E^2 }=g_0·m\]and the gravitational force at a distance h is:\[F_h=G \frac{ M_E·m }{ (R_E+h)^2 }\]. Then we can say:\[F_h=F_E \frac{ R_E^2 }{ (R_E+h)^2 }=m·g_0·\frac{ R_E^2 }{ (R_E+h)^2 }\]The required Energy to move the body from h=0 to infinity is:\[E=m·g_0·R_E^2\int\limits_{0}^{\infty}\frac{ dh }{ (R_E+h)^2 }=m·g_0·R_E^2\left[ \frac{ 1 }{ R_E+h } \right]_\infty^{0}=m·g_0·R_E\]This energy has to equal the kinetic energy: \[m·g_0·R_E=\frac{ 1 }{ 2 }m·v^2\rightarrow v=\sqrt{2·g_0·R_E}\]
  • anonymous
In a nutshell, a velocity of sqrt(2gR) is enough to provide an energy equivalent to the energy required to move the body from the surface of Earth to infinity. It is called escape velocity. Similar reasoning is used to calculate the radius of black holes (Schwarzschild's radius) assuming escape velocity is that of light (c)
  • goformit100
"Welcome to OpenStudy. I can answer your questions or guide you. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( ). Please take a moment to read it."

Looking for something else?

Not the answer you are looking for? Search for more explanations.