Derive the relationship between #"g"# and #"G"#?

Answer 1

#GM=gr^2#
r is the distance between object and the center of earth #r=h+R#
h is height of object from the surface and R is the radius of earth.
and
#g=4/3 piGRrho#

Suppose Earth is a sphere of radius #r#. It has mass #M#. Gravitational Force on the object of mass #m# which is situated at a distance #r# from the center of Earth is #F=(GMm)/r^2# . . .[A]
If the object is free falling from the height h from the surface of earth (or at a distance #r# from the center of earth) it experience the acceleration #g#
According to Newton's second law Force on the object due to acceleration #g# is #F=mg# . . .[B]
Comparing Equation [A] and [B] #cancelmg=(GMcancelm)/r^2# #gr^2=GM# Where #r=h+R# #R="Radius of Earth"# #h="height of object from the surface of earth"#
If #h#<<#R# then we can write #r=R+happroxR# or #gR^2=GM# . . .[C]

(When the object is near the surface of earth so we can neglect the height of object comparing with Radius of Earth)

If average density of earth is #rho# then mass of earth #M="Volume"xx"density"# #M=4/3piR^3rho#
In equation [c] #gR^2=G4/3piR^3rho# #g=(G4/3piR^3rho)/R^2# #g=4/3piGRrho#
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Answer 2

The relationship between "g" (acceleration due to gravity) and "G" (universal gravitational constant) is given by Newton's law of universal gravitation:

[ F = G \frac{m_1 \cdot m_2}{r^2} ]

Where:

  • ( F ) is the gravitational force between two objects,
  • ( G ) is the universal gravitational constant,
  • ( m_1 ) and ( m_2 ) are the masses of the two objects, and
  • ( r ) is the distance between the centers of the two objects.

At the surface of the Earth, ( F ) is the force experienced by an object due to gravity, which is equal to ( m \cdot g ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity. Therefore, at the surface of the Earth:

[ m \cdot g = G \frac{m_{\text{Earth}} \cdot m_{\text{object}}}{R_{\text{Earth}}^2} ]

By rearranging this equation, we can solve for ( g ):

[ g = \frac{G \cdot m_{\text{Earth}}}{R_{\text{Earth}}^2} ]

Where:

  • ( m_{\text{Earth}} ) is the mass of the Earth, and
  • ( R_{\text{Earth}} ) is the radius of the Earth.

Therefore, the relationship between ( g ) and ( G ) is that ( g ) depends on ( G ) as well as the mass and radius of the Earth.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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