Sect. 6-3: Gravity Near Earth’s Surface

g & The Gravitational Constant G

• Obviously, it’s very important to distinguish between G and g– They are obviously very different physical quantities!

G The Universal Gravitational ConstantIt is the same everywhere in the Universe!

G = 6.673 10-11 N∙m2/kg2

It ALWAYS has this value at every location anywhere

g The Acceleration Due to Gravityg = 9.80 m/s2 (approximately!) on Earth’s surface

g varies with location

G vs. g

• Consider an object sitting on Earth’s surface:

mE = mass of Earth (say, known)

RE = radius of Earth (known).

Assume Earth is a uniform, perfect sphere.

m = mass of m (known)

• We defined the Weight of an object of mass m on or near the Earth’s surface as the Gravitational Force on m: FG = mg

• Newton’s Gravitation Law says that the Gravitational force on m is

Fg = G[(mmE)/(rE)2] • Setting these equal gives:

g = 9.8 m/s2

g in terms of Gm

ME

m

ME

Knowing g = 9.8 m/s2 & the radius of the Earth rE, the mass of the Earth can be calculated:

Using the same process, we can

“Weigh” Earth! (Determine it’s mass).

On the surface of the Earth, equate the usual weight of mass m to the Newtonian

Gravitation Force Law form.

Effective Acceleration Due to Gravity• Acceleration due to gravity at a

distance r from Earth’s center.• Write the gravitational force as:

FG = G[(mME)/r2] mg

(effective weight)

g effective acceleration

due to gravity.

SO : g = G (ME)/r2

• If an object is some distance h above the Earth’s surface,

r becomes RE + hAgain, set the gravitational force equal to mg:

G[(mME)/r2] mg This gives:

• This shows that g decreases with increasing altitude• As r , the weight of the object approaches zero• Example 6-4, g on Mt. Everest

2E

E

GMg

R h

Altitude Dependence of g

Altitude Dependence of g

Lubbock, TX: Altitude: h 3300 ft 1100

m g 9.798 m/s2

Mt. Everest: g on Altitude: h 8.8 km

g 9.77 m/s2

2E

E

GMg

R h

Example 6-5: Effect of Earth’s rotation on g

Assuming Earth is perfect sphere, determine how Earth’s rotation affects the value of g at equator compared to its value at pole.

Example 6-5: Effect of Earth’s rotation on g

Assuming Earth is perfect sphere, determine how Earth’s rotation affects the value of g at equator compared to its value at pole is: (Taking towards the Earth’s center as the positive direction)

Newton’s 2nd Law: ∑F = ma = mg – W

W = mg – ma = “effective weight”

At the poles, zero centripetal acceleration: a = 0 W = mg

At the equator, non-zero centripetal acceleration:

aR = [(v2)/(rE)] = W = mg - m[(v2)/(rE)] = mg

g = g - [(v2)/(rE)] = 0.037 m/s2

(v = (2πrE)/T = 4.64 102 m/s, T = 1 day = 8.64 104 s)