- #1
pc2-brazil
- 205
- 3
first of all, we need to say that this question is not for homework.
our doubt is in the process of finding the formula (not the value) of escape velocity in the earth; we want to use the principle of conservation of energy for that purpose.
gravitational potential energy:
we found that there are two formulas for gravitational potential energy:
[tex]U=-\frac{GmM}{R}[/tex] or [tex]U=MgR[/tex], where [tex]G[/tex] is the gravitational constant, [tex]m[/tex] is the mass of the projectile, [tex]M[/tex] is the mass of the earth, [tex]R[/tex] is the radius of the Earth and [tex]g[/tex] is the gravitational acceleration.
kinetic energy:
[tex]K=\frac{mv^{2}}{2}[/tex], where [tex]m[/tex] is the mass of the projectile and [tex]v[/tex] its velocity.
principle of energy conservation:
[tex]K_{0}+U_{0}=K_{1}+U_{1}[/tex]
we found by searching in the internet that the escape velocity of the Earth (or other object) is the minimum velocity needed for a projectile to escape from the Earth's gravitational field (from the surface of the Earth to infinity).
therefore, by the principle of energy conservation:
[tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1}}{2}-\frac{GmM}{r_{1}}[/tex], where [tex]r_{1}=\infty[/tex].
if [tex]r_{1}=\infty[/tex] and the final velocity ([tex]v_{1}[/tex]) is zero:
[tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = 0[/tex]
this derives:
[tex]v_{0}=\sqrt{\frac{GM}{R}}[/tex]
but it's not still our problem.
our problem is the following:
look what happens when we try to substitute the first formula for gravitational potential energy with the second one:
[tex]\frac{mv^{2}_{0}}{2}+MgR = \frac{mv^{2}_{1}}{2}+Mgr_{1}[/tex]
but [tex]r_{1}=\infty[/tex]; it leads to an indetermination:
[tex]Mg\infty[/tex]
if we consider that [tex]Mg\infty[/tex] equals zero, it leads to another impossibility:
[tex]\frac{mv^{2}_{0}}{2}+GmR = 0[/tex]
no more TeX is necessary here to perceive that it will lead to a negative square root.
what are we doing wrong here? we think that the error is in the formulas for gravitational potential energy.
another doubt: what does it mean to set [tex]r_{1}[/tex] as infinity?
thank you in advance for your patience.
Homework Statement
our doubt is in the process of finding the formula (not the value) of escape velocity in the earth; we want to use the principle of conservation of energy for that purpose.
Homework Equations
gravitational potential energy:
we found that there are two formulas for gravitational potential energy:
[tex]U=-\frac{GmM}{R}[/tex] or [tex]U=MgR[/tex], where [tex]G[/tex] is the gravitational constant, [tex]m[/tex] is the mass of the projectile, [tex]M[/tex] is the mass of the earth, [tex]R[/tex] is the radius of the Earth and [tex]g[/tex] is the gravitational acceleration.
kinetic energy:
[tex]K=\frac{mv^{2}}{2}[/tex], where [tex]m[/tex] is the mass of the projectile and [tex]v[/tex] its velocity.
principle of energy conservation:
[tex]K_{0}+U_{0}=K_{1}+U_{1}[/tex]
The Attempt at a Solution
we found by searching in the internet that the escape velocity of the Earth (or other object) is the minimum velocity needed for a projectile to escape from the Earth's gravitational field (from the surface of the Earth to infinity).
therefore, by the principle of energy conservation:
[tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1}}{2}-\frac{GmM}{r_{1}}[/tex], where [tex]r_{1}=\infty[/tex].
if [tex]r_{1}=\infty[/tex] and the final velocity ([tex]v_{1}[/tex]) is zero:
[tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = 0[/tex]
this derives:
[tex]v_{0}=\sqrt{\frac{GM}{R}}[/tex]
but it's not still our problem.
our problem is the following:
look what happens when we try to substitute the first formula for gravitational potential energy with the second one:
[tex]\frac{mv^{2}_{0}}{2}+MgR = \frac{mv^{2}_{1}}{2}+Mgr_{1}[/tex]
but [tex]r_{1}=\infty[/tex]; it leads to an indetermination:
[tex]Mg\infty[/tex]
if we consider that [tex]Mg\infty[/tex] equals zero, it leads to another impossibility:
[tex]\frac{mv^{2}_{0}}{2}+GmR = 0[/tex]
no more TeX is necessary here to perceive that it will lead to a negative square root.
what are we doing wrong here? we think that the error is in the formulas for gravitational potential energy.
another doubt: what does it mean to set [tex]r_{1}[/tex] as infinity?
thank you in advance for your patience.