Determine the gravitational force that you exert on another person 1.20 m away.

Question

Determine the gravitational force that you exert on another person 1.20 m away. Assume that you and the other person are point masses of 66.0 kg each. N During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. What force is exerted by the Sun on the Moon? What force is exerted by the Earth on the Moon? What force is exerted by the Sun on the Earth? Three uniform spheres of masses m_1 = 1.50 kg, m_2 = 4.00 kg, and m_3 = 5.00 kg are placed at the comers of a right triangle (see figure below). Calculate the resultant gravitational force on the object of mass m2, assuming the spheres are isolated from the rest of the Universe. Two identical isolated particles, each of mass 2.50 kg, are separated by a distance of 29.8 cm. What is the magnitude of the gravitational force exerted by one particle on the other? N

Explanation / Answer

(a)

F = G m1 m2/r^2

= 6.67 * 10^-11 (66 kg)^2/(1.2)^2

=2.0 * 10^-7 N

(b)

Earth to Sun distance = (1.496x1011m)

Earth to Moon distance = (3.84x108m)

Moon to Sun distance = 1.496x1011m - 3.84x108m = (1.49216x1011m)

Mass of Earth = (5.98x1024kg)

Mass of Moon = (7.36x1022kg)

Mass of Sun = (1.991x1030kg)Earth to Sun distance = (1.496x1011m)

Earth to Moon distance = (3.84x108m)

Moon to Sun distance = 1.496x1011m - 3.84x108m = (1.49216x1011m)

Mass of Earth = (5.98x1024kg)

Mass of Moon = (7.36x1022kg)

Mass of Sun = (1.991x1030kg)

FSonM = GMsMm/r2

FSonM = (6.673x10-11N-m2/kg2)(1.991x1030kg)(7.36x1022kg)/(1.49216x1011m)2

FSonM = 4.39x1020 N ( toward sun)

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FEonM= GMsMm/r2

FEonM = (6.673x10-11N-m2/kg2) (5.98x1024kg)(7.36x1022kg)/(3.84x108m)2

FEonM = 1.99x1020 N ( Towards earth)

(c)

FSonE = GMsMm/r2

FSonE = (6.673x10-11N-m2/kg2)(5.98x1024kg)(1.991x1030kg)/(1.496x1011m)2

FSonE = 3.55x1022 N ( towards sun)

As per guide liens I worked first two questions please post remaining question sin the next post

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