Iterative Tangents

A curiosity journal of math, physics, programming, astronomy, and more.

Physics

Jumping higher when the sun is overhead

How much higher could someone jump when the sun's gravity is pulling them up? Gravitational acceleration is expressed as a relation of mass, distance, and a gravitational constant:

StartLayout 1st Row 1st Column g 2nd Column equals StartFraction upper G upper M Over d squared EndFraction EndLayout

When the sun is directly overhead, we calculate the net gravitational acceleration by subtracting the sun's pull from the Earth's pull:

StartLayout 1st Row 1st Column g 2nd Column equals g Subscript e a r t h Baseline minus g Subscript s u n Baseline 2nd Row 1st Column Blank 2nd Column equals StartFraction upper G upper M Subscript e a r t h Baseline Over upper R Subscript e a r t h Superscript 2 Baseline EndFraction minus StartFraction upper G upper M Subscript s u n Baseline Over left-parenthesis d Subscript e a r t h minus s u n Baseline minus upper R Subscript e a r t h Baseline right-parenthesis squared EndFraction EndLayout

Earth's gravity pulls someone at the (average) surface with an acceleration of 9.82 m/s². At perihelion, where the sun's gravity is strongest, having the sun directly overhead reduces the net acceleration to 9.81 m/s², about ~(percent-difference g-low g-normal) less than normal.

To find the result in more practical terms, we can calculate a theoretical difference in jumping height using a kinematic equation with no time term:

StartLayout 1st Row 1st Column v Subscript f Superscript 2 2nd Column equals v Subscript i Superscript 2 Baseline plus 2 a d EndLayout

vi is the initial velocity, vf is the final velocity, a is the constant acceleration, and d is distance. For a vertical jump, the maximum height d is reached when vf is 0:

StartLayout 1st Row 1st Column 0 2nd Column equals v Subscript i Superscript 2 Baseline plus 2 a d EndLayout

Solving for initial velocity,

StartLayout 1st Row 1st Column v Subscript i 2nd Column equals StartRoot minus 2 a d EndRoot EndLayout

We can use this vi in the same kinematic equation but with a different value for acceleration, distance now being the unknown:

StartLayout 1st Row 1st Column 0 2nd Column equals v Subscript i Superscript 2 Baseline plus 2 a d 2 2nd Row 1st Column d 2 2nd Column equals minus StartFraction v Subscript i Superscript 2 Baseline Over 2 a 2 EndFraction 3rd Row 1st Column Blank 2nd Column equals minus StartFraction StartRoot minus 2 a d EndRoot squared Over 2 a 2 EndFraction 4th Row 1st Column d 2 2nd Column equals StartFraction a d Over a 2 EndFraction EndLayout

In other words, gravitational acceleration and jump height are inversely proportional. Also notice that the units of a and a₂ cancel out, meaning they don't have to be compatible with the units of d and d₂.

Thus, supposing a 1-foot vertical jump under Earth-only gravity, jump height when the sun is directly overhead is (theoretically) 1.0006 feet, or about 0.2 millimeters higher—a tiny difference, but more visible than I expected.