a body of mass m is projected upward from the earths surface with an initial velocity V0. Take the y-axis to be positive upward, with the origin on the surface of the earth. assuming there is no air resistance, but taking into account the variation of the earths gravitations field with altitude, we obtain that
m (dV/dt) = -(mgR^2) / (y +R)^2
where R is the radius of the earth.
(a) Let V(t) = v(y(t)). Fint a differential equation satisfied by v(y).
(b) find the smallest initial velocity V0 for which the body will not return to earth. this so-called escape velocity. HINT the escape velocity is found by requiring that v(y) remain strictly positive.
Any help would be appreciated!
m (dV/dt) = -(mgR^2) / (y +R)^2
where R is the radius of the earth.
(a) Let V(t) = v(y(t)). Fint a differential equation satisfied by v(y).
(b) find the smallest initial velocity V0 for which the body will not return to earth. this so-called escape velocity. HINT the escape velocity is found by requiring that v(y) remain strictly positive.
Any help would be appreciated!
-
(a) V = v(y), y = y(t)
dV/dt = (dv/dt) = (dv/dy)(dy/dt) = v (dv/dy)
so
m (dV/dt) = -(mgR^2) / (y +R)^2
becomes
m v (dv/dy) = -(mgR^2) / (y +R)^2
(b) solve the differential equation and enforce v > 0
v (dv/dy) = - gR^2 / (y + R)^2
v dv = -(gR^2) dy / (y + R)^2
integrate
(1/2)v^2 = +(gR^2) / (y + R) + C, where C = constant
We need to find C in terms of the initial velocity v0 (note that V0 and v0 are the same thing, I think it is better for bookkeeping to write this as lowercase v0 since our function we relabeled (for some reason) to be lowercase v. We could have just as validly wrote initially V(t) = V(y(t)) [you had written V(t) = v(y(t))]. Case matters in physics and math).
Now, v^2 is strictly positive by definition since v is squared. To find the minimum initial velocity v0, we require minimizing this function v^2. Why? v is the velocity at a position y, escaping means that at the distance at infinity (which is as far away as you get from earth, hence you "escape") you have a velocity of zero, so you need v0 just enough to get you an infinite distance away. Note that in this limit (y -> infinity) the first term on the right hand side goes to zero, v -> 0 by our definition described in the previous sentence, and we are left with the condition C = 0. Another way to think about this is that the form of v above shows some term + C. Minimizing v means adding the smallest amount possible C = 0. The constant C though is expressible in terms of v0, as we will show in a moment
To show that C is expressed in terms of v0, realize that v0 is the initial velocity at y = 0, so for v(y) we have the condition v(0) = v0, thus
(1/2)v^2 = +(gR^2) / (y + R) + C
(1/2)v0^2 = (gR^2) / R + C
C = (1/2)v0^2 - gR
So, minimizing v0 means C = 0
0 = (1/2) v0^2 - gR
v0 = sqrt{2gR} ...............[Ans.]
This result is well-known, most easily people usually get it from balancing kinetic and potential energies at the surface of the earth and the energies at a distance of infinity away from the earth (where the velocity and potential energy is zero, thus it has "escaped' completely from earth's orbit):
1/2 mv0^2 - mgR = 0 ---> v0 = sqrt{2gR}
as we obtained above
dV/dt = (dv/dt) = (dv/dy)(dy/dt) = v (dv/dy)
so
m (dV/dt) = -(mgR^2) / (y +R)^2
becomes
m v (dv/dy) = -(mgR^2) / (y +R)^2
(b) solve the differential equation and enforce v > 0
v (dv/dy) = - gR^2 / (y + R)^2
v dv = -(gR^2) dy / (y + R)^2
integrate
(1/2)v^2 = +(gR^2) / (y + R) + C, where C = constant
We need to find C in terms of the initial velocity v0 (note that V0 and v0 are the same thing, I think it is better for bookkeeping to write this as lowercase v0 since our function we relabeled (for some reason) to be lowercase v. We could have just as validly wrote initially V(t) = V(y(t)) [you had written V(t) = v(y(t))]. Case matters in physics and math).
Now, v^2 is strictly positive by definition since v is squared. To find the minimum initial velocity v0, we require minimizing this function v^2. Why? v is the velocity at a position y, escaping means that at the distance at infinity (which is as far away as you get from earth, hence you "escape") you have a velocity of zero, so you need v0 just enough to get you an infinite distance away. Note that in this limit (y -> infinity) the first term on the right hand side goes to zero, v -> 0 by our definition described in the previous sentence, and we are left with the condition C = 0. Another way to think about this is that the form of v above shows some term + C. Minimizing v means adding the smallest amount possible C = 0. The constant C though is expressible in terms of v0, as we will show in a moment
To show that C is expressed in terms of v0, realize that v0 is the initial velocity at y = 0, so for v(y) we have the condition v(0) = v0, thus
(1/2)v^2 = +(gR^2) / (y + R) + C
(1/2)v0^2 = (gR^2) / R + C
C = (1/2)v0^2 - gR
So, minimizing v0 means C = 0
0 = (1/2) v0^2 - gR
v0 = sqrt{2gR} ...............[Ans.]
This result is well-known, most easily people usually get it from balancing kinetic and potential energies at the surface of the earth and the energies at a distance of infinity away from the earth (where the velocity and potential energy is zero, thus it has "escaped' completely from earth's orbit):
1/2 mv0^2 - mgR = 0 ---> v0 = sqrt{2gR}
as we obtained above