Peoples Physics Book Version-2

3.1. Equivalence between [2] and [3] http://www.ck12.org

raised, its distance from the center of the Earth increases byh. In other words, its new energy is

EG f=

Gm 1 m 2

Rearth+h

[5]

Now we will rewrite equation [5] in the following way:

EG f=

Gm 1 m 2

Rearth

×

Rearth

Rearth+h

which, if we divide the denominator and numerator of the second fraction on the right of the above equation by the
Earth’s radius, gives us:

EG f=

Gm 1 m 2

Rearth

×

1

1 +Rearthh

=

Gm 1 m 2

Rearth

× 1 ×( 1 +

h

Rearth

)−^1

︸ ︷︷ ︸

of the form( 1 +x)−^1

[6]

If we explore the quantities of the type( 1 +x)−^1 , we will see that when|x|1,( 1 +x)−^1 ≈ 1 −x. This kind of
adjustment is called a Taylor approximation. To prove this to yourself, use your calculator to try a bunch of different
numbers and see what the error is.

The ratio ofh, the distance the object was raised, to the radius of the earth is miniscule. So, we can use the theorem
above, remembering that it is this ratio that plays the role ofxabove. This ratio is only small for large objects, such
as planets; and an object’s center of mass may not always be at its geographic center. Both have to be true for our
results to hold. Hence, special case.

Then we find that [6] reduces to

EG f=

Gm 1 m 2

Rearth

×( 1 −

h

Rearth

)[7]

Therefore, we can now rewrite the object’s final potential energy as:

EG f=

Gm 1 m 2

Rearth

−

hGm 1 m 2

Rearth^2

[8]

The last step is to finally calculate the change in potential energy due to the movement. This is straightforward, since

∆E=EGi−EG f= (

Gm 1 m 2

Rearth

)−(

Gm 1 m 2

Rearth

−

hGm 1 m 2

Rearth^2

) =

hGm 1 m 2

Rearth^2

This is the equation we have been looking for. Although it doesn’t look like it, it is completely equivalent to the
formulaEg=mgh. The only variable in this formula is h, everything else — the radius and mass of Earth and G —
are constants. If we rearrange the formula, we see that:

hGm 1 m 2

Rearth^2

=︸︷︷︸m 1

m

×

Gm 2

︸R earth︷︷ 2 ︸

g

×︸︷︷︸h

h

The quantity labeled g, if calculated with the appropriate radius and mass, will give the effective acceleration due to
gravity near the surface. When Earth’s mass and radius are used. the result is 9.8 m/s, but feel free to check.