Iterative Tangents RSS feed

The curiosity journal of a technical creative.

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.