The Earth and a wet, spinning tennis ball are both round, wet, and spins. But how is that water on the surface of the Earth remains there? Unlike water on the surface of the wet, spinning tennis ball.

Can we conclude that the Earth is not actually shaped like a tennis ball? In other words, it is not round.

No, it is not as simple as these flat-earthers are telling you.

Water remains on the surface of the Earth because of gravitational acceleration towards the center of the Earth, caused by the Earth’s mass. There’s actually the same gravitational acceleration in a tennis ball, though not the same amount, not even close. Let’s compare gravitational acceleration of both bodies:

- Gravitational acceleration on the surface of the Earth: 9.82 m/s²
- Gravitational acceleration on the surface of a tennis ball: 0.00000000332 m/s²

On the other hand, there’s also centrifugal acceleration away from the center of the object, caused by rotating motion of both objects. Here’s the comparison:

- Maximum centrifugal acceleration on Earth’s surface: 0.03 m/s²
- Maximum centrifugal acceleration on the surface of a tennis ball that is spinning at 1000 rpm: 376 m/s²

On the surface of the Earth, gravitational acceleration is much greater than the centrifugal acceleration. In contrast, on the surface of a spinning tennis ball that spins at 1000 rpm, centrifugal acceleration is much greater than gravitational acceleration. This is why water on the surface of the Earth remains there, and water on the surface of a spinning tennis ball flies away.

## Calculations

For tennis ball:

- Reference: Tennis Ball – Wikipedia
- Diameter: 2.7 inch
- Mass: 58.5 gram
- Rotational velocity assumption: 1000 rpm (Roger Federer’s backhand can result in 5300 rpm spinning tennis ball).
- Centrifugal acceleration on the surface of a spinning tennis ball: a = ω²r = (((1000 / minutes) * (2 * π))^2) * (2.7 inch / 2) = 376.031928 m / s
^{2} - Gravitational acceleration on the surface of a tennis ball: g = G M / r² = G * 58.5 gram / ( 2.7 inch / 2)^2 = 3.32056743 × 10
^{-9}m / s^{2}

For the Earth:

- Mass: 5.972 × 10^24 kg
- Radius: 6371 km
- Rotational velocity: 1/24 jam
- Gravitational acceleration on the surface of the Earth: g = G M / r² = G * 5.972 × 10^24 kg / (6371 km)^2 = 9.81964974 m / s
^{2} - Centrifugal acceleration on the surface of the Earth: a = ω²r = (((1 / (24 hours)) * (2 * π))^2) * (6371 km) = 0.0336930136 m / s
^{2}