A sensitive gravimeter at a mountian observatory finds that the free-fall acceleration is 6.00 X 10^3 m/s^2 less than that at sea level. What is the observatory's altitude?

g=GMe/(re+altitude)^2

To determine the observatory's altitude, we can utilize the information provided about the difference in free-fall acceleration between the mountaintop and sea level.

Let's start by understanding the relationship between free-fall acceleration and altitude. The acceleration due to gravity decreases as we move away from the Earth's surface. This decrease occurs due to the increasing distance from the center of the Earth and the diminishing gravitational pull.

Now, we can establish the following equation:

acceleration at observatory = acceleration at sea level - 6.00 x 10^3 m/s^2

Next, we need to know the acceleration at sea level, which is a well-known value of approximately 9.8 m/s^2.

Substituting this value into our equation, we get:

acceleration at observatory = 9.8 m/s^2 - 6.00 x 10^3 m/s^2

Calculating the difference, we find:

acceleration at observatory = -5.994 x 10^3 m/s^2

The negative sign indicates that the acceleration at the observatory is lower than at sea level.

Now, we can use the relationship between acceleration and altitude to find the observatory's altitude. The general formula for this relationship is:

acceleration = (GM) / (R + h)^2

Where:
G is the gravitational constant (approximately 6.67430 x 10^-11 m^3/kg/s^2),
M is the mass of the Earth (approximately 5.972 x 10^24 kg),
R is the radius of the Earth (approximately 6.371 x 10^6 m),
and h is the altitude.

Rearranging the formula to solve for h, we have:

h = sqrt((GM) / acceleration) - R

Substituting the known values, we obtain:

h = sqrt((6.67430 x 10^-11 m^3/kg/s^2 * 5.972 x 10^24 kg) / (-5.994 x 10^3 m/s^2)) - 6.371 x 10^6 m

Evaluating this equation will give us the observatory's altitude.