Calculate Acceleration Due To Gravity
What is Acceleration due to Gravity?
Acceleration due to gravity is the acceleration gained by an object due to gravitational forcefulness. Its SI unit of measurement is m/s2. Information technology has both magnitude and management, hence, it'south a vector quantity. Acceleration due to gravity is represented by g. The standard value of grand on the surface of the globe at body of water level is nine.8 m/s2.
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Acceleration due to Gravity – Formula, Unit of measurement and Values
| Acceleration Due to Gravity (thou) | |
| Symbol | thousand |
| Dimensional Formula | M0L1T-two |
| SI Unit | ms-two |
| Formula | g = GM/rtwo |
| Values of chiliad in SI | 9.806 ms-2 |
| Values of g in CGS | 980 cm south-ii |
Table of Content:
- What is Gravity?
- Formula
- g on Earth
- Value of g with Acme
- g with Depth
- g due to Shape of Globe
- 1000 due to Rotation
What is Gravity?
Gravity is the force with which theglobe attracts a torso towards its center. Allow us consider two bodies of masses yarda and mb. Under the awarding of equal forces on two bodies, the force in terms of mass is given by:
mb = granda[aA/aB] this is called an inertial mass of a body.
Nether the gravitational influence on two bodies,
- FA = GMmA/r2,
- FB = GMmB/rii,
- mB = [FB/FA] × mA
⇒ More than on Gravitation:
- Newton's Constabulary of Gravitation
- Gravitational Potential Energy
- Gravitational Field Intensity
The above mass is called a gravitational mass of a body. According to the principle of equivalence, the inertial mass and gravitational mass are identical. We will exist using this while deriving acceleration due to the gravity given beneath.
Let u.s. suppose a torso [test mass (m)] is dropped from a meridian 'h' higher up the surface of the earth [source mass (One thousand)], it begins to move downward with an increment in velocity equally it reaches close to the earth surface.
We know that velocity of an object changes merely nether the action of a force, in this instance, the force is provided by the gravity.
Nether the action of gravitational strength, the trunk begins to accelerate toward the earth'southward middle which is at a altitude 'r' from the test mass.
And so, ma = GMm/r2 (Applying principle of equivalence)
⇒ a = GM/r2 . . . . . . . (1)
The above acceleration is due to the gravitational pull of globe and then we phone call information technology acceleration due to gravity, it does not depend upon the examination mass. Its value well-nigh the surface of the earth is 9.8 ms-2.
Therefore, the acceleration due to gravity (g) is given by = GM/rtwo.
Formula of Dispatch due to Gravity
Forcefulness acting on a torso due to gravity is given by, f = mg
Where f is the force acting on the body, g is the dispatch due to gravity, m is mass of the body.
According to the universal constabulary of gravitation, f = GmM/(r+h)2
Where,
- f = force betwixt two bodies,
- G = universal gravitational abiding (six.67×10-11 Nm2/kg2)
- grand = mass of the object,
- M = mass of the globe,
- r = radius of the globe.
- h = peak at which the torso is from the surface of the earth.
Every bit the superlative (h) is negligibly small-scale compared to the radius of the earth nosotros re-frame the equation as follows,
f = GmM/r2
Now equating both the expressions,
mg = GmM/r2
⇒ thou = GM/rtwo
Therefore, the formula of acceleration due to gravity is given by, g = GM/rii
Note: It depends on the mass and radius of the earth.
This helps u.s. understand the following:
- All bodies feel the same dispatch due to gravity, irrespective of its mass.
- Its value on world depends upon the mass of the earth and not the mass of the object.
Acceleration due to Gravity on the Surface of Globe
World equally assumed to be a compatible solid sphere with a hateful density. We know that,
Density = mass/book
Then, ρ = M/[iv/three πR3]
⇒ Chiliad = ρ × [iv/3 πRthree]
We know that, g = GM/R2.
On substituting the values of M we get,
g = 4/3 [πρRG]
At whatsoever distance 'r' from the heart of the earth
g = iv/3 [πρRG]
The value of acceleration due to gravity 'thousand' is affected by
- Altitude to a higher place the earth's surface.
- Depth below the earth's surface.
- The shape of the earth.
- Rotational motion of the globe.
Variation of chiliad with Height
Acceleration due to Gravity at a summit (h) from the surface of the earth
Consider a test mass (yard) at a peak (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is;
F = GMm/(R+h)2
Where M is the mass of globe and R is the radius of the earth. The acceleration due to gravity at a certain height is 'h' then,
mgh= GMm/(R+h)2
⇒ gh= GM/[R2(1+ h/R)2 ] . . . . . . (two)
The acceleration due to gravity on the surface of the world is given by;
g = GM/R2 . . . . . . . . . (3)
On dividing equation (three) and (2) we get,
yardh= thousand (1+h/R)-2. . . . . . (4)
This is the acceleration due to gravity at a height to a higher place the surface of the earth. Observing the above formula nosotros tin say that the value of g decreases with increase in height of an object and the value of grand becomes zilch at space distance from the earth.
⇒ Check: Kepler'south Laws of Planetary Movement
Approximation Formula:
From Equation (four)
when h << R, the value of 1000 at height 'h' is given by thouh= g/(one – 2h/R)
Variation of g with Depth
Consider a test mass (m) taken to a altitude (d) below the earth's surface, the dispatch due to gravity that point (1000d) is obtained by taking the value of g in terms of density.
On the surface of the earth, the value of g is given by;
g = 4/3 × πρRG
At a distance (d) below the globe'southward surface, the acceleration due to gravity is given by;
yardd = 4/3 × πρ × (R – d) K
On dividing the above equations nosotros get,
gd = one thousand (R – d)/R
- When the depth d = 0, the value of g on the surface of the world gd = g.
- When the depth d = R, the value of k at the centre of the earth md = 0.
Variation of k due to Shape of World
As the earth is an oblate spheroid, its radius well-nigh the equator is more than its radius nearly poles. Since for a source mass, the dispatch due to gravity is inversely proportional to the foursquare of the radius of the globe, it varies with breadth due to the shape of the globe.
gp/ge = R2 due east/R2 p
Where kdue east and gp are the accelerations due to gravity at equator and poles, Reand Rp are the radii of earth near equator and poles, respectively.
From the above equation, it is clear that acceleration due to gravity is more at poles and less at the equator. So if a person moves from the equator to poles his weight decreases as the value of g decreases.
Variation of g due to Rotation of Earth
Consider a exam mass (k) is on a breadth making an angle with the equator. As we have studied, when a body is under rotation every particle in the trunk makes round motions near the axis of rotation. In the present instance, the world is under rotation with a abiding angular velocity ω, then the test mass moves in a circular path of radius 'r' with an angular velocity ω.
This is the case of a non-inertial frame of reference and so there exists a centrifugal force on the test mass (mrω2). Gravity is acting on the exam mass towards the middle of the earth (mg).
As both these forces are acting from the same point these are known as co-initial forces and equally they lie along the same aeroplane they are termed as co-planar forces.
We know from parallelogram constabulary of vectors, if ii coplanar vectors are forming two sides of a parallelogram and so the resultant of those two vectors will always forth the diagonal of the parallelogram.
Applying parallelogram law of vectors we go the magnitude of the apparent value of the gravitational force at the breadth
(mg′)2 = (mg)2 + (mrω2)2 + 2(mg) (mrωii) cos(180 – θ) . . . . . . (1)
Nosotros know 'r' is the radius of the circular path and 'R' is the radius of the world, and so r = Rcosθ.
Substituting r = R cosθ we get,
thousand′ = g – Rω2costwoθ
Where g′ is the credible value of acceleration due to gravity at the breadth due to the rotation of the earth and k is the true value of gravity at the breadth without considering the rotation of the world.
At poles, θ = ninety°⇒ yard' = 1000.
At the equator, θ = 0° ⇒ g′= g – Rωtwo.
Of import Conclusions on Acceleration due to Gravity :
- For an object placed at a height h, the acceleration due to gravity is less equally compared to that placed on the surface.
- Equally depth increases, the value of acceleration due to gravity (grand) falls.
- The value of g is more than at poles and less at the equator.
Gravitation
Gravitational Potential Energy
Frequently Asked Questions on Acceleration due to Gravity
What does the value 9.8 m/s2 for acceleration due to gravity imply?
The value ix.8 m/due south2 for acceleration due to gravity implies that for a freely falling body the velocity changes past 9.8 m/southward every second.
Does mass accept any event on acceleration due to gravity?
The acceleration due to gravity is independent of the mass of the body.
What is the formula to calculate the forcefulness of attraction between two objects?
If chiliad1 and 10002 are the two masses separated by a distance r. According to the universal law of gravitation, the force of attraction between them is
F = G (thousandi yard2/r2)
What is free autumn?
If the object moves only under the influence of gravity information technology is called free fall.
Calculate Acceleration Due To Gravity,
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