Forces and field strengths.
EQUATIONS - Question one-
Gravitational Force ME = 5-97×1024 kg
ME MM
I Newtons Law of gravitation - MM = 7-37×1022 Kg
F= ¥eo 'r? • gravity acts between any 2 objects with mass. P mass = 100kg
F α M, M 2 342×103 km 38×103 km 6=6-67×10-"
E= Far, ALWAYS ATTRACTIVE
4T EO r2
T =
GmiM2 opposite direction (-)
• Force of gravity is an example of an inverse square relationship:
r2
VE-4¥ • C) indicates that force acts in opposite direction to distance.
co r moon → p
gravitational constant = 6-67×10-" ① Earth → P 2
1 Q2 F = 6- 67×10-11 × 5.97×1024×100 F = - 6--67×10-11 ✗ 7.3
PE:*,
r gravitational F α - GM, m² mass of two objects (342×10672 (38×
force between r² F = - 0-34
W = GAVE 2 bodies distance between F = 0 - 34
(WEIGHT) as r increases 2 objects i. {F = 0.34 - 0-34 = ON
F decreases
GM,M2
F = IPS Question two - NEED TO USE VECT
r2 DEFINITION of Newtons Law of gravity: F α plroduct of masses Mm
350×106 m
g: GM The force of attraction between two point masses between 2 points
x
r2 is proportional to the product of the masses of the two
F α 7 EF 80×106 m
points and is inversely proportional to the square of between
Vg:-GI 6.2N
the distance between them. distance? them
r 4×10-3N
M, Mz ME
PE = - GM, Mr
n F Differences between forces, F- = GM/Mz ① Moon → sat
W= mΔVg gravitational and electrostatic? F- = 6- 67×
r2
• gravitational forces are only
r
Constants attractive where as electrostatic
M, and Mz are both attracting eachother forces can be repulsive and h O a 6.2N F = 4 -013
Me = 9.11×10-31 because of newtons 3rd law → one force attractive.
applied between 2 objects. • gravitational deals with masses ② Earth → sa
6=6-67×10-11 electrostatic deals with charges. F = 6-67
SIMILARITIES of 4×010-3N
Electrostatic force CAN BE REPULSIVE
Eo = 8-85×10-12 AND ATTRACTIVE gravitational + electrostatic
F α ⟂ coulombs law ~ point charges • both obey inverse O = tan-' (8) F = 6- 2N
F- α Q, 2 1 82
square law
charge of two objects • both have infinite EF = (6-2)-+ (4×
force between F= 7 1 Q2 range EF = a²+b²
two point charges 4T EO r2 0=0- 037° f
distance between
constant two point charges
= 9.0×109
EQUATIONS - Question one-
Gravitational Force ME = 5-97×1024 kg
ME MM
I Newtons Law of gravitation - MM = 7-37×1022 Kg
F= ¥eo 'r? • gravity acts between any 2 objects with mass. P mass = 100kg
F α M, M 2 342×103 km 38×103 km 6=6-67×10-"
E= Far, ALWAYS ATTRACTIVE
4T EO r2
T =
GmiM2 opposite direction (-)
• Force of gravity is an example of an inverse square relationship:
r2
VE-4¥ • C) indicates that force acts in opposite direction to distance.
co r moon → p
gravitational constant = 6-67×10-" ① Earth → P 2
1 Q2 F = 6- 67×10-11 × 5.97×1024×100 F = - 6--67×10-11 ✗ 7.3
PE:*,
r gravitational F α - GM, m² mass of two objects (342×10672 (38×
force between r² F = - 0-34
W = GAVE 2 bodies distance between F = 0 - 34
(WEIGHT) as r increases 2 objects i. {F = 0.34 - 0-34 = ON
F decreases
GM,M2
F = IPS Question two - NEED TO USE VECT
r2 DEFINITION of Newtons Law of gravity: F α plroduct of masses Mm
350×106 m
g: GM The force of attraction between two point masses between 2 points
x
r2 is proportional to the product of the masses of the two
F α 7 EF 80×106 m
points and is inversely proportional to the square of between
Vg:-GI 6.2N
the distance between them. distance? them
r 4×10-3N
M, Mz ME
PE = - GM, Mr
n F Differences between forces, F- = GM/Mz ① Moon → sat
W= mΔVg gravitational and electrostatic? F- = 6- 67×
r2
• gravitational forces are only
r
Constants attractive where as electrostatic
M, and Mz are both attracting eachother forces can be repulsive and h O a 6.2N F = 4 -013
Me = 9.11×10-31 because of newtons 3rd law → one force attractive.
applied between 2 objects. • gravitational deals with masses ② Earth → sa
6=6-67×10-11 electrostatic deals with charges. F = 6-67
SIMILARITIES of 4×010-3N
Electrostatic force CAN BE REPULSIVE
Eo = 8-85×10-12 AND ATTRACTIVE gravitational + electrostatic
F α ⟂ coulombs law ~ point charges • both obey inverse O = tan-' (8) F = 6- 2N
F- α Q, 2 1 82
square law
charge of two objects • both have infinite EF = (6-2)-+ (4×
force between F= 7 1 Q2 range EF = a²+b²
two point charges 4T EO r2 0=0- 037° f
distance between
constant two point charges
= 9.0×109
