Skip to main content
Search IntMath

Calculating Acceleration Due To Gravity on a Plane

By Kathleen Cantor, 08 Oct 2020

Ever wondered why, when a body is thrown upwards, it comes back down at an increased speed? It is due to the acceleration caused by gravity. Near the earth's surface, there is almost no gravitational force experienced, but it varies at large distances from the earth. Gravity is a force that is experienced between two objects of a particular mass, and it tends to pull them towards the center of the earth.

How to calculate acceleration due to gravity

Acceleration has both magnitude and direction, and thus it is a vector quantity. Acceleration is referred to as the change in velocity per unit time and is denoted by 'a'. The SI unit for acceleration is (m/s2) or newton per kilogram (N/Kg).

This acceleration is dependent on different factors; body mass of an object (m), the distance of the object from the center of mass(r), and the universal gravitational constant (G), which is 6.673*10-11Nm2 Kg2. These factors are related as denoted in the equation; g=GM/r2色.. (i)

, here g represents the acceleration due to gravity.

Factors that affect acceleration due to gravity

Mass is directly proportional to gravity; thus, the heavier the object, the more the gravitational pull it will experience. Gravitational force is inversely proportional to the square of the distance of separation between the bodies. Thus, the larger the distance of separation, the weaker the gravitational pull effect. As two objects are moved away from each other, the gravitational force between them decreases.

Consider to bodies which are 2m away from the center mass and weigh 5 kilograms and 3kilograms, respectively. On substitution using the above equation, it will be seen that the acceleration due to gravity on the 5 kg mass is 1.25G m/s2 while that on the 3 kg mass is 0.75G m/s2. This proves that mass directly affects the gravitational acceleration experienced on a body, i.e, the heavier the mass, the higher the gravitational acceleration felt.

Height is also a factor that affects acceleration due to gravity. Height is inversely proportional to acceleration, which decreases with an increase in height. At an infinite distance from the earth, the value becomes zero. This can be computed in the following equation;

gh=g(1+h/R)-2色. (ii)

gh is the acceleration due to gravity at a particular height, 'h' is the height from the earth's surface and 'R' is the earth's radius.


Assume the moon has a radius of 1.64 * 106m, and has a mass of 7.24 * 1022 Kg. Substituting this, values to the equation (i) above,

I) identify the 'r' and'M' since the universal gravitational constant (G) is constant. In this case,

r = 1.64 * 106 m and M = 7.24 * 1022 Kg

G = 6.673 * 10 -11 Nm2 Kg2

II) Substituting this, values to equation (i) above

Multiply the 'G' and'M' to obtain

= 4.831 * 10 12

III) Divide the above calculated value by r2

r2 =2.689 * 10 12 m2

g = 1.796 ms-2

Therefore, the value of acceleration due to gravity in this case is 1.796 ms-2.

How to calculate acceleration due to gravity on a plane

In calculating the acceleration for a body in motion on a plane, several quantities are put into consideration. These quantities are displacement, speed, and velocity. Displacement is the distance moved towards a particular direction, is denoted by 'S' and the SI unit is meter (m). Speed is the distance covered per unit time, and the SI unit is meters per second (m/s or ms-1). It is a scalar quantity.

Velocity is the change of displacement per unit time and is a vector quantity. The SI unit for velocity is (m/s).

The initial velocity is normally denoted by (u) and the final velocity denoted by (v). In calculating the acceleration of a body, the two velocities must be noted accurately, including the total time taken.

Kinematic Equations Applied

For a uniformly accelerated motion on a plane. The following thee kinematic equations are applied.

v=u+ at色 (iii)

s=ut +1/2 at2 (iv)

v2 =u2 +2as... (v)

It is always observed in our day to day life that leaves on a tree always fall to the ground on their own. Why don't they remain floating or better still, move upwards towards the sky? Or rather, why is it that when a stone is thrown upwards, it falls back to the ground? This phenomenon is due to gravitational force, which causes a free-fall of bodies and objects.

The movement of an object towards the ground involuntarily at an accelerated speed under the sole influence of gravitational force is known as free fall. This acceleration is theoretically estimated to be 9.8 m/s2, but for calculation purposes and for the sake of accuracy, the gravitational force used is rounded off to 10m/s2.

When trying to find the acceleration due to gravity for a free fall, the above kinematic equations are applied, but in this case, acceleration 'a' is replaced with gravitational force 'since it's the force behind the increase in velocity. i.e.

V= u+gt . (vi)

S= ut +1/2 gt 2. (vii)

V2 =u2 +2gs色 (viii)

Vertical Projections

For vertical projections, a body goes against the force of gravity, and thus slowdown is experienced. In this case, the gravitational force is negative.

During a vertical projection, there is a maximum height that can be attained at a particular time. At this height, the final velocity is always zero as the body in motion has to come to a stop before it starts falling back to the ground. Applying the relevant equations, from equation (vi)

V=u+gt安ill be V= u-gt..since v=0 then u=gt and thus t=u/g宇his shows the time taken for an object to attain the maximum height.

From equation (viii)

V2= u2 +2gs安ill be v2 =u2 -2gs in the case of a vertical projection.

Since v=0...then u2=2gHmax.. i.e.'s' is replaced with the maximum height (Hmax). It's still the same thing as both of them represent displacement. Therefore, the maximum height reached can be calculated using the equation Hmax =u2/2g

The time of flight taken by the projectile is twice the time taken to reach the maximum height, thus t=2u/g. In cases here and object returns to its original position, the total displacement equals zero. Thus equation (viii) becomes v2=u2. therefore v=u.


Consider a stone projected vertically upwards with a velocity of 20 m/s.

I) the time taken to reach the maximum height will be;

U=20 g=10 . thus T=20/10 the time taken to reach the maximum height is 2seconds

II) The time of flight is twice the time taken to reach the maximum height, therefore,

4seconds is the time of flight

III) The maximum height reached

Hmax =u2/2g (400/20) therefore the maximum height reached, in this case, is 20m.

IV)The velocity with which the stone lands on the ground, is equal to its initial velocity assuming that there is no air resistance. Thus, the velocity will be 20 m/s.

The projections made could also be horizontal. The path taken during a horizontal projection is known as the trajectory. The maximum horizontal distance covered at a given time I known as range (R). The acceleration is zero at the maximum horizontal distance; thus, from the equation s=ut -1/2 gt2, the range R= ut, and the vertical displacement will be h=1/2gt2.


Gravitational force is responsible for the acceleration observed in an object's movement as it moves towards the ground. A body that undergoes a free fall always accelerates at the rate of 9.8 m/s2.

Be the first to comment below.

Leave a comment

Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

  1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
    `a^2 = sqrt(b^2 + c^2)`
    (See more on ASCIIMath syntax); or
  2. Use simple LaTeX in the following format. Surround your math with \( and \).
    \( \int g dx = \sqrt{\frac{a}{b}} \)
    (This is standard simple LaTeX.)

NOTE: You can mix both types of math entry in your comment.


Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.