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

2. g & The Gravitational Constant G

3. Obviously, it’s very important to distinguish between G and g –They are obviously very different physical quantities! G  The Universal Gravitational Constant It is the same everywhere in the Universe! G = 6.673  10 -11 N∙m 2 /kg 2 It ALWAYS has this value at every location anywhere g  The Acceleration Due to Gravity g = 9.80 m/s 2 (approximately!) on Earth’s surface g varies with location G vs. g

4. Consider an object sitting on Earth’s surface: m E = mass of Earth (say, known) R E = 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: F G = mg Newton’s Gravitation Law says that the Gravitational force on m is F g = G[(mm E )/(r E ) 2 ] Setting these equal gives: g = 9.8 m/s 2 g in terms of G m  MEME

5. MEME Knowing g = 9.8 m/s 2 & the radius of the Earth r E, 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. 

6. Effective Acceleration Due to Gravity Acceleration due to gravity at a distance r from Earth’s center. Write the gravitational force as: F G = G[(mM E )/r 2 ]  mg (effective weight) g  effective acceleration due to gravity. SO : g = G (M E )/r 2

7. If an object is some distance h above the Earth’s surface, r becomes R E + h Again, set the gravitational force equal to mg: G[(mM E )/r 2 ]  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 Altitude Dependence of g

8. Lubbock, TX: Altitude: h  3300 ft  1100 m  g  9.798 m/s 2 Mt. Everest: g on Altitude: h  8.8 km  g  9.77 m/s 2

9. 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.

10. 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 2 nd 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: a R = [(v 2 )/(r E )] = W = mg - m[(v 2 )/(r E )] = mg g = g - [(v 2 )/(r E )] = 0.037 m/s 2 (v = (2πr E )/T = 4.64  10 2 m/s, T = 1 day = 8.64  10 4 s)