Joshua Do
10/26/22
Arsovan 5th
Graph 1/Table 1: Mass is the same, distance increases. (10kg for both masses)
Trial:
Distance
(meter)
Force
(newton)
Trial 1
1m
Trial 2
2m
Trial 3
3m
Trial 4
4m
Trial 5
5m
6.674𝑥10−9 𝑁
1.669𝑥10−9 𝑁
7.42𝑥10−10 𝑁
4.17𝑥10−10 𝑁
2.67𝑥10−10 𝑁
Graph 1 Equation:
6.67𝑥10−9
Force = 𝐷𝑖𝑠 tan 𝑐𝑒 2
Graph 2/Table 2: Mass increases, distance is the same. (4 meters apart)
Trial:
Mass
(m1*m2)
Force
(newton)
Trial 1
Trial 2
Trial 3
Trial 4
100𝑘𝑔2
1000𝑘𝑔2
3000𝑘𝑔2
5000𝑘𝑔2
4.17𝑥10−10 𝑁
4.171𝑥10−9 𝑁
1.2514𝑥10−8 𝑁
2.0857𝑥10−8 𝑁
Graph 2 Equation:
Force = (4.17𝑥10−12 )(𝑚1 ⋅ 𝑚2 )-(7.193𝑥10−14 )
Trial 5
10000𝑘𝑔2
4.1713𝑥10−8 𝑁
How could you use this data to determine Big G, the universal gravitation constant?
Finding Big G (6.67𝑥10−11 ) is as simple as plugging the values gained from the experiment into
𝑚 𝑚
the equation, 𝐹𝑔 = 𝐺 𝑟1 2 2 . The value that would need to be solved for is G, the Gravitational
constant. For example, using data from Graph 2 the equation could be set up that would look
like: 4.17𝑥10−10 𝑁 = 𝐺
100𝑘𝑔2
.
16𝑚
Gravitational Constant.
When solved, G will equal 6.672𝑥10−11
𝑁𝑚 2
.
𝑘𝑔2
This is the Universal