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Murljashka [212]
3 years ago
10

Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple

r’s third law of planetary motion. c. can be used to disprove Kepler’s laws of planetary motion. d. does not apply to Kepler’s laws of planetary motion.
Physics
2 answers:
Finger [1]3 years ago
5 0
Kepler derived his three laws of planetary motion entirely from
observations of the planets and their motions in the sky.

Newton published his law of universal gravitation almost a hundred
years later.  Using some calculus and some analytic geometry, which
any serious sophomore in an engineering college should be able to do,
it can be shown that IF Newton's law of gravitation is correct, then it MUST
lead to Kepler's laws.  Gravity, as Newton described it, must make the planets
in their orbits behave exactly as they do.

This demonstration is a tremendous boost for the work of both Kepler
and Newton.
Nikolay [14]3 years ago
5 0
<h3><u>Answer;</u></h3>

b. can be used to derive Kepler’s third law of planetary motion.

Newton’s law of universal gravitation <u><em>can be used to derive Kepler’s third law of planetary motion</em></u>.

<h3><u>Explanation</u>;</h3>

<u><em>Kepler's</em></u> three laws of planetary motion states that;

  • <em><u>All planets move about the Sun in elliptical orbits, having the Sun as one of the foci. </u></em>
  • <em><u>A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.</u></em>
  • <em><u>The squares of the sidereal periods of the planets are directly proportional to the cubes of their mean distances from the Sun</u></em>

<em><u>Newton's law of universal gravitation</u></em><em><u> states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.</u></em>

<u><em>Newton’s law of universal gravitation can be used to derive Kepler’s third law of planetary motion.</em></u>

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio
LUCKY_DIMON [66]

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29274.93096 m/s

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Explanation:

r_p = Distance at perihelion = 1.471\times 10^{11}\ m

r_a = Distance at aphelion = 1.521\times 10^{11}\ m

v_p = Velocity at perihelion = 3.027\times 10^{4}\ m/s

v_a = Velocity at aphelion

m = Mass of the Earth =  5.98 × 10²⁴ kg

M = Mass of Sun = 1.9889\times 10^{30}\ kg

Here, the angular momentum is conserved

L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J

Kinetic energy at perihelion is 2.73966\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times  10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}

Potential energy at perihelion is -5.39323\times 10^{33}\ J

K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J

Kinetic energy at aphelion is 2.56249\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}

Potential energy at aphelion is -5.21594\times 10^{33}\ J

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