# Newton’s Law of Gravitation Problems and Solutions

**Problem#1**

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 – m

^{2})/r

^{2}

The force will be maximum, when

dF/dm = 0

G(M – 2m)/r

^{2}= 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 10

^{12}N?

Answer:

Known:

Mass of particle, m

_{1}= 5.2 kg and m

_{2}= 2.4 kg

Large force of interaction between the two particles is, F = 2.3 x 10

^{12}N

Using the the gravitational force law:

F = Gm

_{1}m

_{2}/r

^{2}

F = 2.3 x 10

^{12}N = (6.67x10⁻¹¹ Nm²/kg²)(5.2 kg)(2.4 kg)/r

^{2}

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 10

^{30}kg

Mass of Earth, M

_{E}= 5.98 x 10

^{24}kg

Radius of sun, r

_{S}= 1.50 x 10

^{11}m and

Radius of Earth, r

_{E}= 3.82 x 10

^{8}m

the force of interaction between the moon and the Sun is

F

_{Sun}= GM

_{S}M

_{m}/r

_{S}

^{2}

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

F

_{Earth}= GM

_{E}M

_{m}/r

_{S}

^{2}

The ratio is

*F*

_{Sun}/

*F*

_{Earth}= M

_{s}r

_{E}

^{2}/[M

_{E}r

_{S}

^{2}]

*F*

_{Sun}/

*F*

_{Earth}= [1.99 x 10

^{30}kg x (3.82 x 10

^{8}m)

^{2}]/[5.98 x 10

^{24}kg x (1.50 x 10

^{11}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 10

^{8}m and Earth’s radius is 6.37 x 10

^{6}m.

Answer:

Distance between the moon and the Earth r

_{ME}= 3.82 x 10

^{8}m

Distance from the surface of the Earth and the moon r

_{E}= 6.37 x 10

^{6}m

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

The gravitational force pulls on you before changing noted by F

_{1}, with

F

_{1}= GM

_{m}m/(r

_{ME}+ r

_{E})

^{2}(*)

The gravitational force pulls on you after changing noted by F

_{2}, with

F

_{2}= GM

_{m}m/(r

_{ME}– r

_{E})

^{2}(**)

The ration between F

_{1}and F

_{2}, then subtract the result by 1 (since the ratio without any change will equal 1), so that

F

_{2}/F

_{1}= (r

_{ME}+ r

_{E})

^{2}/(r

_{ME}– r

_{E})

^{2}

F

_{2}/F

_{1 }– 1 = [(r

_{ME}+ r

_{E})

^{2}/(r

_{ME}– r

_{E})

^{2}] – 1

F

_{2}/F

_{1 }– 1 = [(3.82 x 10

^{8}m + 6.37 x 10

^{6}m)

^{2}/(3.82 x 10

^{8}m – 6.37 x 10

^{6}m)

^{2}] – 1

F

_{2}/F

_{1 }– 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

F

_{2}– F

_{1}= GM

_{m}m/(r

_{ME}– r

_{E})

^{2 }– GM

_{m}m/(r

_{ME}+ r

_{E})

^{2}

F

_{2}– F

_{1}= F

_{1}[F

_{2}/F

_{1}– 1]

F

_{2}– F

_{1}= 0.069GM

_{m}m/(r

_{ME}+ r

_{E})

^{2}

Your weight on the Earth F

_{E}

F

_{E}= GM

_{E}m/r

_{E}

^{2}

With M

_{E}massa of the Earth

Then, the ratio,

(F

_{2}– F

_{1})/F

_{E}= 0.069M

_{m}r

_{E}

^{2}/[M

_{E}(r

_{ME}+ r

_{E})

^{2}]

(F

_{2}– F

_{1})/F

_{E}= 0.069 x [7.36 x 10

^{22 }kg/5.98 x 10

^{24}kg][6.37 x 10

^{6}m/(6.37 x 10

^{6}m + 3.82 x 10

^{8}m)]

^{2}

(F

_{2}– F

_{1})/F

_{E}= 2.28 x 10

^{-7}x 100% = 2.28 x 10

^{-5}%

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