Review an Object of Mass M Is Dropped From Height H Above a Planet of Mass M and Radius R
What is Acceleration due to Gravity?
Acceleration due to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is m/southward2. Information technology has both magnitude and direction, hence, it's a vector quantity. Acceleration due to gravity is represented by 1000. The standard value of g on the surface of the world at sea level is 9.8 yard/due south2.
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Acceleration due to Gravity – Formula, Unit and Values
| Acceleration Due to Gravity (g) | |
| Symbol | chiliad |
| Dimensional Formula | K0L1T-two |
| SI Unit of measurement | ms-2 |
| Formula | g = GM/r2 |
| Values of g in SI | ix.806 ms-two |
| Values of g in CGS | 980 cm southward-ii |
Table of Content:
- What is Gravity?
- Formula
- g on Globe
- Value of k with Height
- g with Depth
- k due to Shape of Earth
- grand due to Rotation
What is Gravity?
Gravity is the strength with which theearth attracts a body towards its centre. Let united states consider two bodies of masses ma and mb. Under the application of equal forces on two bodies, the force in terms of mass is given by:
thoub = ma[aA/aB] this is called an inertial mass of a body.
Under the gravitational influence on two bodies,
- FA = GMmA/rtwo,
- FB = GMmB/r2,
- mB = [FB/FA] × mA
⇒ More than on Gravitation:
- Newton'due south Constabulary of Gravitation
- Gravitational Potential Free energy
- Gravitational Field Intensity
The higher up mass is called a gravitational mass of a torso. According to the principle of equivalence, the inertial mass and gravitational mass are identical. We will be using this while deriving acceleration due to the gravity given below.
Let us suppose a body [test mass (g)] is dropped from a height 'h' above the surface of the earth [source mass (1000)], it begins to motion downwards with an increase in velocity every bit it reaches close to the globe surface.
We know that velocity of an object changes only under the action of a force, in this case, the force is provided past the gravity.
Nether the activeness of gravitational forcefulness, the body begins to accelerate toward the earth's centre which is at a distance 'r' from the test mass.
Then, ma = GMm/r2 (Applying principle of equivalence)
⇒ a = GM/r2 . . . . . . . (1)
The above dispatch is due to the gravitational pull of globe so we call information technology acceleration due to gravity, it does not depend upon the test mass. Its value near the surface of the earth is 9.8 ms-ii.
Therefore, the acceleration due to gravity (one thousand) is given by = GM/rtwo.
Formula of Dispatch due to Gravity
Force acting on a body due to gravity is given by, f = mg
Where f is the force acting on the body, 1000 is the acceleration due to gravity, k is mass of the body.
According to the universal law of gravitation, f = GmM/(r+h)2
Where,
- f = force between two bodies,
- K = universal gravitational constant (6.67×10-11 Nm2/kg2)
- thou = mass of the object,
- M = mass of the earth,
- r = radius of the earth.
- h = height at which the body is from the surface of the earth.
As the height (h) is negligibly modest compared to the radius of the earth we re-frame the equation equally follows,
f = GmM/r2
Now equating both the expressions,
mg = GmM/rtwo
⇒ one thousand = GM/rtwo
Therefore, the formula of acceleration due to gravity is given by, g = GM/rtwo
Note: It depends on the mass and radius of the globe.
This helps us understand the following:
- All bodies experience the aforementioned acceleration due to gravity, irrespective of its mass.
- Its value on earth depends upon the mass of the earth and not the mass of the object.
Dispatch due to Gravity on the Surface of Earth
Earth as assumed to be a uniform solid sphere with a mean density. We know that,
Density = mass/volume
Then, ρ = M/[four/three πR3]
⇒ M = ρ × [4/iii πR3]
We know that, g = GM/Rii.
On substituting the values of M we get,
g = 4/3 [πρRG]
At any distance 'r' from the heart of the globe
thousand = 4/3 [πρRG]
The value of acceleration due to gravity 'g' is afflicted by
- Altitude above the world'due south surface.
- Depth below the world'due south surface.
- The shape of the globe.
- Rotational move of the earth.
Variation of k with Height
Acceleration due to Gravity at a height (h) from the surface of the globe
Consider a test mass (k) at a tiptop (h) from the surface of the globe. Now, the force interim 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 sure height is 'h' so,
mgh= GMm/(R+h)ii
⇒ grandh= GM/[R2(one+ h/R)two ] . . . . . . (ii)
The acceleration due to gravity on the surface of the world is given by;
g = GM/Rii . . . . . . . . . (iii)
On dividing equation (3) and (2) nosotros get,
gh= g (i+h/R)-2. . . . . . (4)
This is the acceleration due to gravity at a meridian above the surface of the earth. Observing the above formula we can say that the value of thou decreases with increase in superlative of an object and the value of yard becomes zero at infinite distance from the globe.
⇒ Check: Kepler'due south Laws of Planetary Move
Approximation Formula:
From Equation (4)
when h << R, the value of g at height 'h' is given past gh= thou/(1 – 2h/R)
Variation of 1000 with Depth
Consider a test mass (k) taken to a distance (d) beneath the earth'due south surface, the acceleration due to gravity that bespeak (thousandd) is obtained by taking the value of chiliad in terms of density.
On the surface of the earth, the value of k is given past;
g = four/3 × πρRG
At a distance (d) below the world'due south surface, the acceleration due to gravity is given past;
gd = 4/iii × πρ × (R – d) G
On dividing the above equations we get,
grandd = thousand (R – d)/R
- When the depth d = 0, the value of g on the surface of the earth gd = chiliad.
- When the depth d = R, the value of g at the center of the earth yardd = 0.
Variation of chiliad due to Shape of Earth
As the world is an oblate spheroid, its radius nearly the equator is more than its radius near poles. Since for a source mass, the dispatch due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.
chiliadp/ge = Rtwo e/Rtwo p
Where ge and one thousandp are the accelerations due to gravity at equator and poles, Reastwardand Rp are the radii of world virtually 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 k decreases.
Variation of g due to Rotation of Earth
Consider a exam mass (yard) is on a latitude making an angle with the equator. Equally we accept studied, when a body is under rotation every particle in the body makes circular motions about the axis of rotation. In the nowadays case, the globe is nether rotation with a constant angular velocity ω, so the test mass moves in a circular path of radius 'r' with an angular velocity ω.
This is the instance of a non-inertial frame of reference so there exists a centrifugal forcefulness on the test mass (mrω2). Gravity is interim on the test mass towards the center of the globe (mg).
Every bit both these forces are acting from the same point these are known equally co-initial forces and every bit they lie along the same airplane they are termed equally co-planar forces.
Nosotros know from parallelogram law of vectors, if two coplanar vectors are forming 2 sides of a parallelogram then the resultant of those two vectors volition always along the diagonal of the parallelogram.
Applying parallelogram police of vectors nosotros get the magnitude of the apparent value of the gravitational force at the breadth
(mg′)two = (mg)2 + (mrωii)2 + 2(mg) (mrω2) cos(180 – θ) . . . . . . (1)
We know 'r' is the radius of the circular path and 'R' is the radius of the earth, then r = Rcosθ.
Substituting r = R cosθ nosotros get,
g′ = thousand – Rωiicosiiθ
Where one thousand′ is the apparent value of acceleration due to gravity at the breadth due to the rotation of the earth and m is the truthful value of gravity at the latitude without considering the rotation of the globe.
At poles, θ = 90°⇒ 1000' = g.
At the equator, θ = 0° ⇒ g′= 1000 – Rωtwo.
Of import Conclusions on Acceleration due to Gravity :
- For an object placed at a acme h, the acceleration due to gravity is less every bit compared to that placed on the surface.
- Every bit depth increases, the value of acceleration due to gravity (yard) falls.
- The value of g is more at poles and less at the equator.
Source: https://byjus.com/jee/acceleration-due-to-gravity/
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