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.

Answer:
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 M is split into two parts, m and M – m, which are then separated by a certain distance. What ratio m/M maximizes the magnitude of the gravitational force between the parts?

Answer:
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?

Answer:
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 F_Sun/F_Earth of these two forces? (The average Sun–Moon distance is equal to the Sun–Earth distance.)

Answer:
Known:
Mass of sun, M_sun = 1.99 x 10³⁰ kg
Mass of Earth, M_E = 5.98 x 10²⁴ kg
Radius of sun, r_S = 1.50 x 10¹¹ m and
Radius of Earth, r_E = 3.82 x 10⁸ m

the force of interaction between the moon and the Sun is

F_Sun = GM_SM_m/r_S²

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

F_Earth = GM_EM_m/r_S²

The ratio is

F_Sun/F_Earth = M_sr_E²/[M_Er_S²]

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

F_Sun/F_Earth = 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.

Answer:
Distance between the moon and the Earth r_ME = 3.82 x 10⁸ m
Distance from the surface of the Earth and the moon r_E = 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 = GM_mm/(r_ME + r_E)² (*)
The gravitational force pulls on you after changing noted by F₂, with

F₂ = GM_mm/(r_ME – r_E)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 = (r_ME + r_E)²/(r_ME – r_E)²

F₂/F₁ – 1 = [(r_ME + r_E)²/(r_ME – r_E)²] – 1    

F₂/F₁ – 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/F₁ – 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 = GM_mm/(r_ME – r_E)² – GM_mm/(r_ME + r_E)² 

F₂ – F₁ = F₁[F₂/F₁ – 1]

F₂ – F₁ = 0.069GM_mm/(r_ME + r_E)²

Your weight on the Earth F_E

F_E = GM_Em/r_E²

With M_E massa of the Earth

Then, the ratio,

(F₂ – F₁)/F_E = 0.069M_mr_E²/[M_E(r_ME + r_E)²]

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