€

Fg =GmMR2

F ∝ mF ∝ MF ∝ mM

F ∝1R2

6.67 x 10-11

N-m2/kg2

mg

€

Fg =GmMR2

→ mg =GmMR2

→ g =GMR2

“PHLYZICS” Newton’s Universal Law of Gravitation

• All objects are attracted to each other. In other words, all objects exert attractive forces on

each other. • The larger an object’s mass, the larger the attractive force it exerts. • As you move away from an object, the force decreases, BUT NOT LINEARLY. The

force variation is inversely proportional to the square of the distance from the center. • Newton’s Universal Law, in equation form, says that the force between two objects is

• G has a VERY small value, approximately ________________, with units of

__________.

• Newton’s “ULOG” is a vector law!!!

• Since the force due to gravity on an objects is known as the object’s weight, and since

weight equals ______, then

Where: m = mass of smaller object M = mass of larger object R = distance between objects G = Newton’s Universal Gravitational Constant

At each labeled point above, find the weight of an 80 kg person, the acceleration due to gravity felt by the person, and the force of attraction between tbe person and the earth. Each point is exactly 1 “earth’s radius” away from the previous point, except for point E, which is 10,000 m above the earth’s surface.

rearth = 6.4 x6 m

Mearth = 6 x24 kg

A

B

C

D

E

Newton’s Universal Law of Gravitation (ULOG)Worksheet (SOLUTIONS)

€

Fg =GmMR2

=(6.67E −11m 3

s2 )(5kg)(10kg)(.3m)2

= 3.71E − 8N

€

W = mg → m =W /g =100N /9.8 ms2 =10.204kg

Fg =GmMR2

=(6.67E −11m 3

s2 )(10.204kg)(6E24kg)(6.38E6m + 6.4E6m)2

= 25N

€

g =GMR2

=(6.67E −11m 3

s2 )(6E24kg)(1,000,000m + 6.4E6m)2

= 7.3 ms2

€

Fg =Gm1m2

R2

if m1 is tripled, then Fg =Gm1(3m2)

R2 =3Gm1m2

R2 = 3Fg

if R is halved, then Fg =Gm1m2

R2( )

2 =Gm1m2

R2

4( )=

4Gm1m2

R2 = 4Fg

if both happen simultaneously, then Fg =3⋅ 4Gm1m2

R2 =12Gm1m2

R2 =12Fg

#1 #2 #3 #4 #5

€

Fg =Gm1m2

R2

In order to quadruple the force, you must halve the distance.

Fg =Gm1m2

R2 =Gm1m2

(10m)2 =Gm1m2

100m2 =14N → Gm1m2 =1400N ⋅ m2

Fg =Gm1m2

R2 = 56N

1400N ⋅ m2

R2 = 56N → R2 =1400N ⋅ m2

56N= 25m2 → R = 5m

€

Fg =Gm1m2

R2

if m1 is doubled, then Fg =Gm1(2m2)

R2 =2Gm1m2

R2 = 2Fg

if R is quadrupled, then Fg =Gm1m2

4R( )2 =Gm1m2

16R2 =1

16Fg

if both happen simultaneously, then

Fg =2Gm1m2

16R2 =1Gm1m2

8R2 =18Fg =

18

(100N) =12.5N

If the moon’s mass were to double, then both the centripetal force on the moon would double and the gravitational force between it and the earth would double. This would cause the centripetal force to remain balanced with the gravitational force. Therefore, the orbital radius would not need to adjust, and the moon would continue in its orbit. HOWEVER, since the gravitational force would double, the moon would pull harder on the earth, thus causing (among other issues) much higher tides, possibly of the Tsunami variety!!!

€

Fg =GmMR2

→ G =Fgr

2

mM=N ⋅ m2

kg2=(kg⋅ m

s2)⋅ m2

kg2=

m3

kg⋅ s2

#6 #7 #8 #9

€

Fg =GmMR2

=(6.67E −11 Nm 2

kg 2 )(1.67E − 27kg)(9.11E − 31kg)(5E −11m)2

= 4E − 47N

€

gearth's surface =GMR2 =

GMre

2

If re is halved, g will be quadrupled.If M is divided by 1/8, then g will also be reduced by 1/8th.

Therefore, gmar's surface = 4⋅ 18⋅GMre

2 =12GMre

2 =12gearth' s surface =

12

(9.8 ms2 ) = 4.9 m

s2

€

g =GMR2

=GMre2 =

(6.67E −11 Nm 2

kg 2 )(5.67E26kg)(6.3E7m)2

= 9.5 ms2

Fg = mg = (9.5 ms2 )(60kg) = 570N

€

Fg =Gm1m2

R2or W = mg (both should yield the same answer)

Fg =G(5.98E24)(50)(6.38E6)2

= 489.95N W = mg = (50)(9.8) = 490N

€

Fg =Gm1m2

R2

if m1 is doubled, then Fg =Gm1(2m2)

R2 =2Gm1m2

R2 = 2Fg

if R is tripled, then Fg =Gm1m2

3R( )2 =Gm1m2

9R2 =19Fg

if both happen simultaneously, then Fg =2Gm1m2

9R2 =29Fg =

29

(36N) = 8N

€

g =GMJ

rJ2 =

(6.67E −11 Nm 2

kg 2 )(1.9E27kg)(7.2E7m)2

= 24 ms2

#10 #11 #12 #13 #14 #15

When you go into a hole, mass is both below you and above you. Therefore, as you go down, you are not pulled only downward (by the large mass below you), but also upward (by the smaller, but increasing, mass above you).

His weight will decrease proportionally to the square of the distance.

€

Fg =GmMR2 =

(6.67E −11m 3

s2 )(1kg)(6E24kg)(6.4E6m + 6.4E6m)2 = 2.45N

Fg = .5W = .5mg =GmMR2 → .5g =

GMR2

R =GM.5g

= 9,037,337.5m

the distance from the surface is R − rE = 2,637,337m

in terms of earth radii, R = 2,637,337m /6.4E6m = .41rE

#16 #17 #18 #19

€

Fg =GmMR2 =

(6.67E −11m 3

s2 )(1kg)(6E24kg)(6.4E6m + 6.4E6m)2 = 2.45N

Fg = .01W = .01mg =GmMR2 → .01g =

GMR2 →

R =GM.01g

= 63,903,626.4m

the distance from the surface is R − rE = 57,503,626.4m

in terms of earth radii, R = 57,503,626.4m /6.4E6m = 9rE

€

Fg =GmEMS

R2=(6.67E −11m 3

s2 )(1.99E30kg)(6E24kg)(1.5E11m)2

= 3.54E22N

€

Fg =GmMR2

=(6.67E −11m 3

s2 )(55kg)(70kg)(10m)2

= 2.56E − 9N

Since both forces exert equal but opposite forces on each other, and since F = ma applies to both, the acceleration of the larger mass will be half as much as the

acceleration of the smaller mass.

F = (2m)abig F = ma small

(2m) abig = ma small 2 abig = a small abig = ½ a small

€

Fg =GmMR2

F = Fg (←) + Fg (→)

F =G(1500kg)(1400kg)

(7m)2(←) +

G(1400kg)(800kg)(10m)2

(→)

= 2.859E − 6N(←) + 7.470E − 7N(→)= 2.1E − 6N(←)

#20 #21 #22 #23 #24a

€

Fg =GmMR2

F = Fg (↓) + Fg (→)

F =G(1500kg)(1300kg)

(7m)2(↓) +

G(1500kg)(800kg)(10m)2

(→)

= 2.654E − 6N(↓) + 8E − 7N(→)= 2.8E − 6N(right 73o down)

€

Fg =GmMR2

F = Fg (↓) + Fg (→)

F =G(1500kg)(1300kg)

(7m)2(↓) +

G(1500kg)(800kg)(12m)2

⎛

⎝ ⎜

⎞

⎠ ⎟ cos50o(↓) +

G(1500kg)(800kg)(12m)2

⎛

⎝ ⎜

⎞

⎠ ⎟ sin50o(→)

= 2.654E − 6N(↓) + 3.57E − 7(↓) + 4.258E − 7N(→)=# E − 6N(↓) + 4.258E − 7N(→)= 3E − 6N (right 82o down)

#24b #24c