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Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $

Square root and integrating:

$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $

This gives the solution

$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $

However, my problem with this is that it seems to imply that the velocity scales:

$ \dot{x} \sim t^{-1/3} ,$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).

This should be straightforward, but I'm missing something?

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $$

Square root and integrating:

$$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $$

This gives the solution

$$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $$

However, my problem with this is that it seems to imply that the velocity scales:

$$ \dot{x} \sim t^{-1/3} ,$$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).

This should be straightforward, but I'm missing something?

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $

Square root and integrating:

$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $

This gives the solution

$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $

However, my problem with this is that it seems to imply that the velocity scales:

$ \dot{x} \sim t^{-1/3} ,$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $

Square root and integrating:

$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $

This gives the solution

$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $

However, my problem with this is that it seems to imply that the velocity scales:

$ \dot{x} \sim t^{-1/3} ,$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).

This should be straightforward, but I'm missing something?

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $

Square root and integrating:

$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $

This gives the solution

$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $

However, my problem with this is that it seems to imply that the velocity scales:

$ \dot{x} \sim t^{-1/3} ,$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity?

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $

Square root and integrating:

$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $

This gives the solution

$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $

However, my problem with this is that it seems to imply that the velocity scales:

$ \dot{x} \sim t^{-1/3} ,$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).