Escape Velocity, with varying acceleration as a(y)

[Chris T.]

Homework Statement

Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations

a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution

My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.

 

[kishlaysingh]

acceleration due to gravity is a function of position

 

[vela]

Homework Statement

Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations

a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution

My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.

You can use the chain rule to write
$$a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v \frac{dv}{dy}.$$

 

[Dick]

Homework Statement

Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations

a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution

My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.

Using conservation of energy is the easiest way to approach this if you know how that works.

 

[Chris T.]

You can use the chain rule to write
$$a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v \frac{dv}{dy}.$$

So using the substitution a = vdv/dy and the given equation for acceleration, then separating variables, I should be able to integrate both sides and solve for velocity with respect to y. Let me try that.

 

[Chris T.]

Using conservation of energy is the easiest way to approach this if you know how that works.

After integration, I find v^2 = 2g(R^2)/(R+y) + C. Is it valid to say that C is zero because v is zero as y goes to infinity?