metre per second squared

SI coherent derived unit

Name	Symbol	Quantity	Base units
metre per second squared	m s⁻²	acceleration	m s⁻²
The metre per second squared, symbol m s⁻², is the SI coherent derived unit of acceleration. An object with a constant acceleration of one metre per second squared experiences an increase in velocity of one metre per second over a period of one second.
Definition			c Δν_Cs
$1 \mspace{4mu} \text{m} \mspace{4mu} \text{s}^{-2} \mspace{6mu} = \dfrac{1}{(299 \mspace{4mu} 792 \mspace{4mu} 458)(9 \mspace{4mu} 192 \mspace{4mu} 631 \mspace{4mu} 770)} \mspace{6mu} c \mspace{4mu} \Delta \nu _{Cs}\\ \\ \\ 1 \mspace{4mu} \text{m} \mspace{4mu} \text{s}^{-2} \mspace{6mu} \approx 3.628 \mspace{4mu} 602 \mspace{4mu} 815 \mspace{4mu} 210 \mspace{4mu} 469 \mspace{4mu} 913 \times10^{-19} \mspace{6mu} c \mspace{4mu} \Delta \nu _{Cs}$

Acceleration is a vector quantity – it has both magnitude and direction. It is the rate of change of velocity of an object with respect to time. An object’s acceleration is the net result of all forces acting on the object.

Newton’s second law of motion

Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

The law can also be stated in terms of an object’s acceleration. For a constant-mass system, the net force, F, applied to an object of mass, m, is directly proportional to the product of the object’s mass, m, and its acceleration, a.

$F \propto m \mspace{2mu} a$

Using SI coherent units, the proportionality constant is 1. Thus:

$F = m \mspace{2mu} a$

where:

F is the net force applied in newtons, symbol N,
m is the mass of the object in kilograms, symbol kg,
a is the object’s acceleration in metres per second squared, symbol m s^-2.

Newton’s law of universal gravitation

Newton’s law of universal gravitation states that every point mass attracts every other point mass by a force acting along the line intersecting the two points. The force is directly proportional to the product of the two masses, and inversely proportional to the square of the distance between them.

$F \propto \dfrac{m_1 \mspace{2mu} m_2} {r^2}$

Using SI coherent units, the proportionality constant is the gravitational constant, G. Thus:

$F = G \mspace{2mu} \dfrac{m_1 \mspace{2mu} m_2} {r^2}$

where:

F is the gravitational force acting between the two objects in newtons, symbol N,
G is the gravitational constant, equal to 6.674 30(15) × 10^-11 N m² kg^-2,
m₁ and m₂ are the masses of the two objects in kilograms, symbol kg,
r is the distance between the two objects in metres, symbol m.

Acceleration due to Earth’s gravity

Re-arranging the above equation, and substituting unit mass, m, for m₁, and the mass of the Earth, M, for m₂, gives an expression for the gravitational force per unit mass at the surface of the Earth:

$\dfrac{F}{m} = \dfrac{G \mspace{2mu} M} {r^2}$

where:

F is gravitational force, in newtons, symbol N, acting between a unit mass, m, and the mass of the Earth, M, in kilograms,
G is the gravitational constant, equal to 6.674 30(15) × 10^-11 kg^-1 m³ s^-2,
M is the mass of the Earth, equal to approximately 5.972 × 10²⁴ kg,
r is the radius of the Earth, equal to approximately 6.371 × 10⁶ m.

If the symbol g is used to represent acceleration due to Earth’s gravity, then from Newton’s Second Law,

$g = \dfrac{F}{m}$

Acceleration due to Earth’s gravity, g, can then be calculated as follows:

$g \approx \dfrac{(6.674 \mspace{4mu} 30 \times10^{-11} \mspace{4mu} \text{kg}^{-1} \mspace{2mu} \text{m}^3 \mspace{2mu} \text{s}^{-2}) \times (5.972 \times10^{24} \mspace{4mu} \text{kg})} {(6.371 \times10^6 \mspace{4mu} \text{m})^2}\\ \\ \\ g \approx 9.8 \mspace{6mu} \text{m} \mspace{4mu} \text{s}^{-2}$