Question

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance ℎ above the surface of the Earth, ????ℎ . Express the equation in terms of the radius ???? of the Earth, ???? , and ℎ .

Answer 1

g(h) = g ( 1 - 2(h/R) )

*At first order on h/R*

Step-by-step explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

`F = G (m_1m_2)/(r^2)`

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

`F = m_2a`

Therefore, we can see that

`a(r) = G (m_1)/(r^2)`

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

`a(R+h) \approx a(R) + h (da(r))/(dr)_(r=R)`

a(R) is actually the constant acceleration at sea level

and

`a(R) =G (m_1)/(R^2) \\ (da(r))/(dr)_(r=R) = -2 G(m_1)/(R^3)`

Therefore:

`a(R+h) \approx G(m_1)/(R^2) -2G(m_1)/(R^2) (h)/(R) = g(1-2(h)/(R))`

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

`a(R+h) \approx a(R) + h (da(r))/(dr)_(r=R) + (h^2)/(2!) (d^2a(r))/(dr^2)_(r=R)`

Answer 2

The acceleration of Earth's gravity at any distance ' R ' from its center is

g = G m/r²

G = the gravitational constant

m = Earth's mass

r = the distance from the Earth's center

As long as you're walking around on the Earth's surface, your distance from its center is the Earth's radius ... big 'R' ... and the acceleration of gravity is (G m/R²) no matter where on Earth you are.

But if you get into a hot-air balloon or an airplane, and raise yourself up to 'h' off of the surface, then you're farther from the center of the Earth. Now your distance from the center is (R + h), and the acceleration of gravity up there where you are is (G m/(R+h)²) ... that's less than it is down on the surface.

So there's the equation we're looking for.

Acceleration of gravity at any altitude 'h' = G m / (R+h)²