SME
Grade 11 Science · Problem type
Universal gravitation
Use this for “How hard do Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg), 3.84 × 10⁸ m apart, pull on each other? What if they were twice as far?”
Diagram
F
=
Gm₁m₂/
r
2
F
=
6.67 ×
10
-11
×
5.97 ×
10
24
×
7.35 ×
10
22
/384,000,
000
2
F
=
1.99 ×
10
20
N
Each mass pulls the other just as hard (Newton’s third law).
Twice the distance, a quarter of the pull.
This is an example. Type your own values to start a new problem.
F (N)
=
6.674
×
10
−11
×
m₁ (kg)
×
m₂ (kg)
(
r (m)
)
2
Formulas
F
=
6.674 ×
10
-11
×
m₁
×
m₂
/
r
2
1.99 ×
10
20
=
6.674 ×
10
-11
×
(5.97
×
10
24
)
×
(7.35
×
10
22
)/
(3.84 ×
10
8
)
2
Assumptions
•
Every two masses pull on each other: F = Gm₁m₂/r², G = 6.674 × 10⁻¹¹ N·m²/kg².
•
r is the distance between their centers. Twice as far gives a quarter of the pull.
•
The pulls on the two are equal and opposite, whatever their masses.
Step-by-step
WE KNOW
m₁ = 5.97 × 10²⁴ kg, m₂ = 7.35 × 10²² kg, r = 3.84 × 10⁸ m
FIND
pull of gravity (F)
Step 1 · Find pull of gravity (F)
Use F = 6.674 × 10⁻¹¹ × m₁ × m₂/r²
G
=
6.674 ×
10
-11
N·
m
2
/kg²:
multiply
G
by
both
masses
and
divide
by
the
distance
squared.
F
=
6.674 ×
10
-11
×
m₁
×
m₂
/(
r
2
)
F
=
6.674 ×
10
-11
×
(5.97
×
10
24
)
×
(7.35
×
10
22
)/(
(3.84 ×
10
8
)
2
)
m₁
×
m₂
=
(5.97
×
10
24
)
×
(7.35
×
10
22
)
=
4.388 ×
10
47
kg²
r
2
=
(3.84 ×
10
8
)
2
=
1.4746 ×
10
17
m
2
F
=
6.674 ×
10
-11
×
(4.388
×
10
47
)/(1.4746
×
10
17
)
F
=
1.99 ×
10
20
N
Check
Put the numbers back in. Both sides should match.
1.99 ×
10
20
≈
6.674 ×
10
-11
×
(5.97
×
10
24
)
×
(7.35
×
10
22
)/
(3.84 ×
10
8
)
2
✓
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