• Example: On the summit of Mount Everest (8,848 meters above sea level), \( g \) is slightly lower than at sea level.
  
  • At sea level, \( g \approx 9.80665 \, \text{m/s}^2 \).
    
  • At the summit of Everest, \( g \approx 9.764 \, \text{m/s}^2 \).
    

The decrease is due to being farther from Earth’s center, where gravitational force weakens.

—

• The Kola Superdeep Borehole in Russia reaches about 12.3 km below Earth’s surface.
  
• As you go deeper into Earth, \( g \) decreases because you are moving closer to Earth’s core and the mass above you starts to counteract some of the gravitational pull.
  
  • At a depth of 12 km, \( g \) is only slightly reduced, by about 0.03% compared to sea level.
    

—

• Latitude: \( g \) is higher at the poles (9.832 m/s²) than at the equator (9.780 m/s²) due to Earth’s oblate shape and rotational effects.
  
• Local geology: Variations in the density of Earth’s crust (e.g., mountains, valleys, or mineral deposits) cause small local fluctuations.
  

—

• Sea level (average): \( g \approx 9.80665 \, \text{m/s}^2 \)
  
• Mount Everest (8.8 km): \( g \approx 9.764 \, \text{m/s}^2 \) (decrease of ~0.4%)
  
• Kola Superdeep Borehole (12.3 km depth): \( g \approx 9.803 \, \text{m/s}^2 \) (decrease of ~0.03%)
  
• Poles vs Equator: Difference of ~0.5%.
  

These variations are small but measurable with precise instruments.

Can we calculate how high from the surface of
the earth we need to go up for g to be half of
its value at see level?

Yes, we can calculate the height above the Earth’s surface where \( g \) becomes half its value at sea level using the formula for gravitational acceleration:

\[ g = G \frac{M}{r^2} \]

where:

• \( G \) is the gravitational constant,
  
• \( M \) is Earth’s mass,
  
• \( r \) is the distance from Earth’s center.
  

At sea level, \( r = R \), where \( R \) is Earth’s radius (\( R \approx 6,371 \, \text{km} \)), and \( g_0 \) (gravitational acceleration at sea level) is proportional to \( \frac{1}{R^2} \).

We want \( g \) to be half of \( g_0 \), so:

\[ \frac{g}{g_0} = \frac{1}{2} = \frac{R^2}{(R + h)^2} \]

Simplifying for \( h \) (the height above Earth’s surface):

\[ \sqrt{2} R = R + h \]

\[ h = (\sqrt{2} - 1) R \]

Substitute \( R = 6,371 \, \text{km} \):

\[ h = (\sqrt{2} - 1) \cdot 6,371 \, \text{km} \]

\[ h \approx 0.414 \cdot 6,371 \, \text{km} \]

\[ h \approx 2,638 \, \text{km} \]

You need to go approximately 2,638 km above Earth’s surface for \( g \) to be half of its value at sea level.

If we carry 1 kg from the surface of the earth
to h=2,638 km, how much would 1 kg weigh in
that altitude?

To find the weight of a 1 kg object at an altitude of \( h = 2,638 \, \text{km} \), we use the formula for gravitational acceleration:

\[ g_h = G \frac{M}{(R + h)^2} \]

where:

• \( g_h \) is the gravitational acceleration at height \( h \),
  
• \( g_0 \) is the gravitational acceleration at the surface (\( 9.80665 \, \text{m/s}^2 \)),
  
• \( R \) is Earth’s radius (\( 6,371 \, \text{km} \)).
  

At \( h = 2,638 \, \text{km} \), we already calculated that \( g_h = \frac{1}{2} g_0 \). So:

\[ g_h = 0.5 \times 9.80665 = 4.9033 \, \text{m/s}^2 \]

The weight of the object at this altitude is:

\[ \text{Weight} = m \cdot g_h \]

For \( m = 1 \, \text{kg} \):

\[ \text{Weight} = 1 \cdot 4.9033 = 4.9033 \, \text{N} \]

At an altitude of 2,638 km, a 1 kg object would weigh approximately 4.90 N.

I don’t understand something.
So my text says (from Newton’s Principia)
that “in regions where the accelerative gravity [g]
is less by a factor of two, the weight of a body
that is smaller by a factor of two will be less
by a factor of four.”
So is this confirmed by your last computation?

Let’s carefully analyze the statement from Newton’s Principia to see if it aligns with the computations:

Newton’s statement essentially describes how gravitational acceleration (\( g \)) and weight relate. It says:

• If \( g \) is reduced by a factor of 2 (\( g_h = 0.5 \cdot g_0 \)),
  
• Then for a body whose weight is also halved (e.g., \( 1 \, \text{kg} \to 0.5 \, \text{kg} \)),
  
• The new weight would be reduced further, by a factor of 4 relative to the original.
  

This implies:
\[ \text{New weight} = \text{Original weight} \times \frac{1}{4} \]

—

From the computations:

• At \( h = 2,638 \, \text{km} \), \( g_h = 0.5 \cdot g_0 \).
  
• A \( 1 \, \text{kg} \) object at \( h \) weighed \( 4.90 \, \text{N} \), which is half of its sea-level weight (\( 9.81 \, \text{N} \)).
  

Now, if we reduce the mass of the object by half (from \( 1 \, \text{kg} \) to \( 0.5 \, \text{kg} \)):

• The weight of the object at \( h \) would then be:
  

\[ \text{New weight} = 0.5 \, \text{kg} \times 4.9033 \, \text{m/s}^2 = 2.45 \, \text{N} \]

This new weight (\( 2.45 \, \text{N} \)) is:
\[ \text{New weight compared to sea level} = \frac{2.45}{9.81} = \frac{1}{4} \]

—

Yes, the statement from Newton’s Principia is confirmed by the computation:

• Reducing \( g \) by a factor of 2 reduces the weight by a factor of 2.
  
• Reducing the mass by an additional factor of 2 causes the final weight to be reduced by a total factor of 4 relative to the original at sea level.