Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on

Question

3 years ago

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Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on

ly known force a planet exerts on Earth is gravitational.
(a) Calculate the gravitational force Father exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).
(b) Calculate the force F Jupiter on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29 x 10^11 m away.
(c) How does the force of Jupiter on the baby compare to the force of the father on the baby Father / OF Jupiter

Physics

1 answer:

astraxan [27] · Answer 1 · 2022-04-27T21:58:44+03:00

Answer: (a) $F=7(10)^{-7}N$

(b) $F=1.344(10)^{-6}N$

(c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father $m_{1}=100kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between them $r=0.2m$ , the force $F_{F}$ exerted by the father is:

$F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}$ (2)

$F_{F}=0.0000007N=7(10)^{-7}N$ (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter $m_{J}=1.898(10)^{27}kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between Jupiter and Earth (where the baby is) $r_{E}=6.29(10)^{11}m$ , the force $F_{J}$ exerted by the Jupiter is:

$F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}$ (4)

$F_{J}=0.000001344N=1.344(10)^{-6}N$ (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

$F_{J}=0.000001344N=1.344(10)^{-6}N$ and $F_{F}=0.0000007N=7(10)^{-7}N$ we can see $F_{J}$ is greater than $F_{F}$ . However, the difference is quite small as well as the force exerted on the baby.