# Newton’s Law of Gravitation Problems and Solutions

Problem#1

Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them.

Known:
Mass of each ball, m = 10 kg
The distance between them, r = 10 cm = 0.10 m
The universal constant of gravitation, G = 6.67x10⁻¹¹ Nm²/kg²
From the universal law of gravitation the gravitational force of attraction between the balls,

F = Gm.m/r²

F = (6.67x10⁻¹¹ Nm²/kg²)(10 kg)(10 kg)/(0.10 m)²

F = 6.67x10⁻⁷ N

Problem#2
A mass is split into two parts, and – m, which are then separated by a certain distance. What ratio m/maximizes the magnitude of the gravitational force between the parts?

From the universal law of gravitation the gravitational force of attraction between the balls,

F = Gm(M – m)/r²

F = G(Mm – m2)/r2

The force will be maximum, when

dF/dm = 0

G(M – 2m)/r2 = 0

M – 2m = 0

m/M = ½

Problem#3
What must the separation be between a 5.2 kg particle and a 2.4 kg particle for their gravitational attraction to have amagnitude of 2.3 x 1012 N?

Known:
Mass of particle, m1 = 5.2 kg and m2 = 2.4 kg
Large force of interaction between the two particles is, F = 2.3 x 1012 N

Using the the gravitational force law:

F = Gm1m2/r2

F = 2.3 x 1012 N = (6.67x10⁻¹¹ Nm²/kg²)(5.2 kg)(2.4 kg)/r2

r = 19 m

Problem#4
The Sun and Earth each exert a gravitational force on the Moon.What is the ratio FSun/FEarth of these two forces? (The average Sun–Moon distance is equal to the Sun–Earth distance.)

Known:
Mass of sun, Msun = 1.99 x 1030 kg
Mass of Earth, ME = 5.98 x 1024 kg
Radius of sun, rS = 1.50 x 1011 m and
Radius of Earth, rE = 3.82 x 108 m

the force of interaction between the moon and the Sun is

FSun = GMSMm/rS2

the force of interaction between the moon and the Earth sun is

FEarth = GMEMm/rS2

The ratio is

FSun/FEarth = MsrE2/[MErS2]

FSun/FEarth = [1.99 x 1030 kg x (3.82 x 108 m)2]/[5.98 x 1024 kg x (1.50 x 1011 m)2]

FSun/FEarth = 2.16

Problem#5
Moon effect. Some people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon’s gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth–Moon (center-to-center) distance is  3.82 x 108 m and Earth’s radius is 6.37 x 106 m.

Distance between the moon and the Earth rME = 3.82 x 108 m
Distance from the surface of the Earth and the moon rE = 6.37 x 106 m

(a) Percent does the Moon’s gravitational pull on you increase,

The gravitational force pulls on you before changing noted by F1, with

F1 = GMmm/(rME + rE)2 (*)
The gravitational force pulls on you after changing noted by F2, with

F2 = GMmm/(rME – rE)2    (**)

The ration between F1 and F2, then subtract the result by 1 (since the ratio without any change will equal 1), so that

F2/F1 = (rME + rE)2/(rME – rE)2

F2/F– 1 = [(rME + rE)2/(rME – rE)2] – 1

F2/F– 1 = [(3.82 x 108 m + 6.37 x 106 m)2/(3.82 x 108 m – 6.37 x 106 m)2] – 1

F2/F– 1 = 0.069

Percentage = 6.9%

(b) your weight (as measured on a scale) decrease?

The percentage of the decreased weight is the ratio between the difference in gravitational pull on you and your wight on the Earth.

Finding the change in gravitational pull on you

F2 – F1 = GMmm/(rME – rE)– GMmm/(rME + rE)2

F2 – F1 = F1[F2/F1 – 1]

F2 – F1 = 0.069GMmm/(rME + rE)2

Your weight on the Earth FE

FE = GMEm/rE2

With ME massa of the Earth

Then, the ratio,

(F2 – F1)/FE = 0.069MmrE2/[ME(rME + rE)2]

(F2 – F1)/FE = 0.069 x [7.36 x 1022 kg/5.98 x 1024 kg][6.37 x 106 m/(6.37 x 106 m + 3.82 x 108 m)]2

(F2 – F1)/FE = 2.28 x 10-7 x 100% = 2.28 x 10-5%