Why is it hard to accelerate forwards in a car under low gravity? I saw a physics simulation where a dude was driving a car under normal vs low gravity, and it was notably harder to accelerate forwards/change direction under lower gravity. any explanation for this?
 A: Assuming that the car is a wheel-driven vehicle, then the horizontal force that accelerates the car is actually the friction between the car's tyres and the road surface. The maximum friction force is proportional to the vertical normal force exerted by the road on the car, which is in turn (on a level road) equal to the car's weight.
In lower gravity the car's weight is reduced so the maximum friction force between the car and the road surface is reduced. This reduces the maximum acceleration of the car.
In higher gravity the car's weight is increased so the maximum friction force between the car and the road surface is increased, and the car can in theory accelerate more quickly (remember that although the car weighs more in high gravity, its mass is unchanged). However, there is another limiting factor on the car's maximum acceleration - the maximum torque that the engine can produce on the car's wheels. This will depend on the power of the engine and the ratio of the car's lowest gear.
A: This is because the typical car accelerates by exerting a force on the ground. The reaction force the ground exerts on the car then propels the car forward. In a low gravity situation, the car is not in solid contact with the ground, there is no reaction force, and hence it cannot accelerate.
Note this doesn't apply if the car uses a jet engine, which works by expelling air backwards. In that situation the car will still be able to accelerate under low gravity.
A: I am assuming "notably harder to accelerate forwards" means that the car loses traction sooner at low gravity vs normal gravity. The reason that is the case is traction is lost when the static friction force, which is responsible for acceleration, reaches the maximum possible static friction force sooner, which is
$$F_{f-max}=\mu_{s}mg$$
Where $mg$ is the portion of the weight of the car on the drive wheel and $\mu_{s}$ is the coefficient of static friction between the tire and road. Assuming $m$ and $\mu_s$ are constant, $F_{f-max}$ is lower if $g$ is lower.
Hope this helps.
