at what altitude above the earths surface is the acceleration of gravity equal to g/143?

What is Acceleration due to Gravity?

Acceleration due to gravity is the dispatch gained past an object due to gravitational force. Its SI unit is g/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 thou on the surface of the world at sea level is 9.8 m/southwardⁱⁱ.

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Acceleration due to Gravity – Formula, Unit and Values

Acceleration Due to Gravity (g)
Symbol	m
Dimensional Formula	Chiliad0L1T-ii
SI Unit	ms-2
Formula	g = GM/rtwo
Values of thou in SI	9.806 ms-two
Values of k in CGS	980 cm southward-2

Table of Content:

• What is Gravity?
• Formula
• g on Earth
• Value of g with Peak
• g with Depth
• k due to Shape of Earth
• g due to Rotation

What is Gravity?

Gravity is the force with which theearth attracts a body towards its eye. Let us consider two bodies of masses m_a and k_b. Under the application of equal forces on two bodies, the strength in terms of mass is given past:

m_b = thousand_a[a_A/a_B] this is chosen an inertial mass of a body.

Under the gravitational influence on 2 bodies,

• F_A = GMm_A/r^two,
• F_B = GMm_B/r²,
• one thousand_B = [F_B/F_A] × thousand_A

⇒ More than on Gravitation:

• Newton'south Law of Gravitation
• Gravitational Potential Energy
• Gravitational Field Intensity

The in a higher place mass is called a gravitational mass of a body. Co-ordinate 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 torso [test mass (m)] is dropped from a height 'h' in a higher place the surface of the globe [source mass (M)], it begins to move downward with an increase in velocity as it reaches close to the world surface.

We know that velocity of an object changes only nether the action of a force, in this case, the strength is provided by the gravity.

Nether the activeness of gravitational forcefulness, the body begins to advance toward the earth's centre which is at a distance 'r' from the exam mass.

Then, ma = GMm/rⁱⁱ (Applying principle of equivalence)

⇒ a = GM/r^two . . . . . . . (1)

The to a higher place acceleration is due to the gravitational pull of earth and so we call information technology acceleration due to gravity, it does not depend upon the test mass. Its value nearly the surface of the globe is 9.8 ms^-2.

Therefore, the acceleration due to gravity (g) is given past = GM/r².

Formula of Acceleration due to Gravity

Force acting on a body due to gravity is given by, f = mg

Where f is the force acting on the torso, g is the dispatch due to gravity, m is mass of the body.

Co-ordinate to the universal law of gravitation, f = GmM/(r+h)²

Where,

• f = force between two bodies,
• G = universal gravitational constant (half dozen.67×10^-11 Nm²/kg²)
• m = mass of the object,
• 1000 = mass of the world,
• r = radius of the earth.
• h = pinnacle at which the body is from the surface of the earth.

As the top (h) is negligibly small compared to the radius of the earth we re-frame the equation every bit follows,

f = GmM/r^two

Now equating both the expressions,

mg = GmM/r²

⇒ m = GM/r²

Therefore, the formula of acceleration due to gravity is given by, g = GM/r²

Note: Information technology depends on the mass and radius of the globe.

This helps us sympathise 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 globe and non the mass of the object.

Dispatch due to Gravity on the Surface of World

Earth as assumed to exist a uniform solid sphere with a mean density. We know that,

Density = mass/volume

So, ρ = Yard/[4/3 πR³]

⇒ M = ρ × [4/3 πR³]

Nosotros know that, g = GM/R².

On substituting the values of M we get,

g = iv/3 [πρRG]

At whatever distance 'r' from the centre of the world

one thousand = 4/3 [πρRG]

The value of  dispatch due to gravity 'g' is affected by

• Altitude above the earth's surface.
• Depth beneath the earth'due south surface.
• The shape of the globe.
• Rotational motility of the earth.

Variation of g with Meridian

Dispatch due to Gravity at a elevation (h) from the surface of the earth

Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the exam 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' and then,

mg_h= GMm/(R+h)^two

⇒ 1000_h= GM/[Rⁱⁱ(i+ h/R)² ] . . . . . . (2)

The acceleration due to gravity on the surface of the globe is given by;

one thousand = GM/Rⁱⁱ . . . . . . . . . (3)

On dividing equation (3) and (2) we become,

g_h= g (ane+h/R)^-two. . . . . . (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 g decreases with increase in height of an object and the value of g becomes zero at space distance from the globe.

⇒ Check: Kepler's Laws of Planetary Motility

Approximation Formula:

From Equation (four)

when h << R, the value of one thousand at height 'h' is given by g_h= g/(1 – 2h/R)

Variation of g with Depth

Consider a test mass (m) taken to a distance (d) beneath the globe's surface, the acceleration due to gravity that point (g_d) 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 = four/three × πρRG

At a distance (d) below the earth's surface, the acceleration due to gravity is given by;

g_d = iv/3 × πρ × (R – d) G

On dividing the above equations we become,

yard_d = chiliad (R – d)/R

• When the depth d = 0, the value of chiliad on the surface of the earth g_d = g.
• When the depth d = R, the value of g at the center of the globe yard_d = 0.

Variation of g due to Shape of Earth

As the earth is an oblate spheroid, its radius virtually the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the globe, it varies with breadth due to the shape of the earth.

thou_p/g_e = R² _east/R² _p

Where thousand_e and g_p are the accelerations due to gravity at equator and poles, R_eand R_p are the radii of earth about equator and poles, respectively.

From the higher up equation, information technology 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 nosotros have studied, when a torso is under rotation every particle in the body makes circular motions about the axis of rotation. In the nowadays example, 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 athwart velocity ω.

This is the case of a non-inertial frame of reference then there exists a centrifugal forcefulness on the test mass (mrωⁱⁱ). Gravity is acting on the exam mass towards the centre of the globe (mg).

As both these forces are acting from the same signal these are known as co-initial forces and every bit they lie along the aforementioned aeroplane they are termed as co-planar forces.

We know from parallelogram law of vectors, if two coplanar vectors are forming two sides of a parallelogram then the resultant of those two vectors will e'er forth the diagonal of the parallelogram.

Applying parallelogram law of vectors we get the magnitude of the credible value of the gravitational force at the latitude

(mg′)² = (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 we become,

g′ = yard – Rω²cos^twoθ

Where g′ is the apparent value of acceleration due to gravity at the latitude due to the rotation of the earth and g is the true value of gravity at the latitude without considering the rotation of the earth.

At poles, θ = 90°⇒ g' = g.

At the equator, θ = 0° ⇒ k′= g – Rω².

Important Conclusions on Acceleration due to Gravity :

• For an object placed at a height h, the dispatch due to gravity is less as compared to that placed on the surface.
• Every bit depth increases, the value of acceleration due to gravity (g) falls.
• The value of m is more at poles and less at the equator.

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