what is the magnitude and direction of the acceleration due to gravity
What is Acceleration due to Gravity?
Dispatch due to gravity is the dispatch gained past an object due to gravitational forcefulness. Its SI unit is yard/s2. It has both magnitude and direction, hence, information technology's a vector quantity. Acceleration due to gravity is represented by g. The standard value of m on the surface of the earth at sea level is ix.eight thousand/s2.
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Acceleration due to Gravity – Formula, Unit and Values
| Dispatch Due to Gravity (grand) | |
| Symbol | k |
| Dimensional Formula | M0L1T-2 |
| SI Unit | ms-ii |
| Formula | chiliad = GM/r2 |
| Values of g in SI | 9.806 ms-2 |
| Values of 1000 in CGS | 980 cm s-2 |
Table of Content:
- What is Gravity?
- Formula
- chiliad on Earth
- Value of chiliad with Peak
- thou with Depth
- chiliad due to Shape of Earth
- yard due to Rotation
What is Gravity?
Gravity is the force with which theworld attracts a torso towards its centre. Let united states consider ii bodies of masses yarda and mb. Under the application of equal forces on two bodies, the strength in terms of mass is given by:
mb = ma[aA/aB] this is chosen an inertial mass of a body.
Under the gravitational influence on two bodies,
- FA = GMmA/r2,
- FB = GMmB/rii,
- one thousandB = [FB/FA] × mA
⇒ More on Gravitation:
- Newton's Law of Gravitation
- Gravitational Potential Energy
- Gravitational Field Intensity
The to a higher place mass is called a gravitational mass of a trunk. 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 u.s.a. suppose a torso [exam mass (m)] is dropped from a height 'h' above the surface of the earth [source mass (K)], information technology begins to movement down with an increase in velocity every bit it reaches close to the earth surface.
We know that velocity of an object changes only under the action of a force, in this case, the forcefulness is provided by the gravity.
Nether the action of gravitational force, the body begins to advance toward the earth'south center which is at a distance 'r' from the test mass.
Then, ma = GMm/rii (Applying principle of equivalence)
⇒ a = GM/r2 . . . . . . . (1)
The above acceleration is due to the gravitational pull of world so we telephone call it dispatch due to gravity, it does not depend upon the examination mass. Its value near the surface of the earth is ix.eight ms-2.
Therefore, the dispatch due to gravity (g) is given by = GM/rtwo.
Formula of Acceleration due to Gravity
Force acting on a body due to gravity is given by, f = mg
Where f is the force interim on the body, k is the acceleration due to gravity, m is mass of the body.
According to the universal law of gravitation, f = GmM/(r+h)2
Where,
- f = force between two bodies,
- G = universal gravitational constant (6.67×x-11 Nm2/kgtwo)
- m = mass of the object,
- M = mass of the earth,
- r = radius of the world.
- h = height at which the body is from the surface of the earth.
As the height (h) is negligibly pocket-size compared to the radius of the earth we re-frame the equation equally follows,
f = GmM/r2
At present equating both the expressions,
mg = GmM/rii
⇒ g = GM/r2
Therefore, the formula of acceleration due to gravity is given by, g = GM/r2
Note: It depends on the mass and radius of the globe.
This helps united states understand the following:
- All bodies experience the same 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.
Acceleration due to Gravity on the Surface of Globe
World as causeless to exist a uniform solid sphere with a mean density. We know that,
Density = mass/volume
So, ρ = 1000/[4/3 πRthree]
⇒ M = ρ × [4/three πR3]
Nosotros know that, g = GM/R2.
On substituting the values of 1000 we get,
yard = 4/3 [πρRG]
At any distance 'r' from the centre of the earth
k = 4/3 [πρRG]
The value of acceleration due to gravity 'g' is affected by
- Altitude to a higher place the earth's surface.
- Depth beneath the world's surface.
- The shape of the earth.
- Rotational motion of the globe.
Variation of m with Height
Dispatch due to Gravity at a height (h) from the surface of the earth
Consider a test mass (1000) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is;
F = GMm/(R+h)two
Where M is the mass of earth and R is the radius of the earth. The acceleration due to gravity at a sure height is 'h' then,
mgh= GMm/(R+h)2
⇒ thouh= GM/[Rii(1+ h/R)2 ] . . . . . . (two)
The acceleration due to gravity on the surface of the world is given by;
g = GM/R2 . . . . . . . . . (three)
On dividing equation (3) and (2) we become,
thouh= m (one+h/R)-two. . . . . . (4)
This is the acceleration due to gravity at a summit to a higher place the surface of the earth. Observing the above formula we tin can say that the value of chiliad decreases with increment in height of an object and the value of g becomes zero at space distance from the earth.
⇒ Check: Kepler'south Laws of Planetary Motion
Approximation Formula:
From Equation (4)
when h << R, the value of yard at height 'h' is given by gh= g/(1 – 2h/R)
Variation of m with Depth
Consider a exam mass (m) taken to a distance (d) below the earth's surface, the acceleration due to gravity that point (gd) is obtained by taking the value of g in terms of density.
On the surface of the world, the value of g is given by;
g = 4/3 × πρRG
At a altitude (d) below the world'south surface, the acceleration due to gravity is given past;
yardd = 4/3 × πρ × (R – d) Chiliad
On dividing the above equations we get,
one thousandd = g (R – d)/R
- When the depth d = 0, the value of g on the surface of the earth gd = g.
- When the depth d = R, the value of one thousand at the center of the world grandd = 0.
Variation of thousand due to Shape of World
As the earth is an oblate spheroid, its radius nigh the equator is more than than its radius nigh poles. Since for a source mass, the acceleration 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 globe.
gp/thousande = R2 due east/R2 p
Where me and gp are the accelerations due to gravity at equator and poles, Rdue eastand Rp are the radii of world 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. Then 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 test mass (m) is on a breadth making an bending with the equator. As we take studied, when a torso is under rotation every particle in the torso makes circular motions about the centrality of rotation. In the nowadays case, the globe is under rotation with a constant angular velocity ω, then the exam 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 forcefulness on the exam mass (mrω2). Gravity is interim on the test mass towards the middle of the globe (mg).
Every bit both these forces are interim from the same signal these are known every bit co-initial forces and as they lie along the same airplane they are termed as co-planar forces.
We know from parallelogram law of vectors, if ii coplanar vectors are forming 2 sides of a parallelogram and then the resultant of those ii vectors will ever along the diagonal of the parallelogram.
Applying parallelogram law of vectors we become the magnitude of the apparent value of the gravitational force at the latitude
(mg′)2 = (mg)2 + (mrωtwo)2 + 2(mg) (mrωii) 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 we get,
1000′ = g – Rωtwocos2θ
Where g′ is the apparent value of acceleration due to gravity at the breadth due to the rotation of the world and g is the truthful value of gravity at the breadth without considering the rotation of the globe.
At poles, θ = 90°⇒ g' = g.
At the equator, θ = 0° ⇒ g′= g – Rωii.
Of import Conclusions on Acceleration due to Gravity :
- For an object placed at a height h, the acceleration due to gravity is less as compared to that placed on the surface.
- Equally depth increases, the value of dispatch due to gravity (g) falls.
- The value of m is more than at poles and less at the equator.
Source: https://byjus.com/jee/acceleration-due-to-gravity/
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