#1 You know that the weight is
GMm/r^2 = 100
Replace M by 7M and r by 2r, and you have the new weight of
G(7M)m/(2r)^2 = 7/4 GMm/r^2 = 175
If an object which weighs 100 lbs on the Earth's surface were placed on a planet with 2 times the radius of the Earth and with 7 times the Earth's mass, how much would that object weigh? Enter answer to nearest 0.1 lbs.
The term "void ratio" means the ratio of the volume of empty space to volume of occupied space. It is a measure of how empty a volume is. Within stellar systems, the average separation between planets is roughly 10 to the 15 cm and the average planetary radius is typically 10 to the 15 cm. The void ratio for this stellar system is roughly 10 to what power: [Hint: Remember that volume is proportional to the cube of the linear size]
Given that the acceleration of gravity at the Earth's surface is about 980 cm per sec per sec, the centripetal acceleration [v squared over r] of an artificial satellite in a circular orbit with a radius 3.9 times that of the Earth's radius would be about _____ cm per sec per sec. (Hint: F = m a, and the force of gravity gets weaker as the square of the distance, so how much weaker is gravity at 3.9 times the Earth's Radius than it is at the surface of the Earth?)
If the distance between the Earth and the Sun were increased by a factor of 4.56, by what factor would the strength of the force between them change? [Hint: Use Newton's Law of Universal Gravitation, and give your answer to 2 decimal places only]
If a certain force accelerates an object of mass 27 Kg at 50 m/s/s, what acceleration in m/s/s would the same force produce on another object of mass 27? Enter answer to at least one decimal place to the right of the decimal point.
Show work please I need to know how to do it
2 answers
#2
Huh? The planetary radius is equal to the separation? Do the planets touch, or is the separation between planetary surfaces?
If so, then a cube with side 5r contains about 9 planets of radius r. So, the void ratio would be
((5r)^3 - 9*4/3 π*r^3)/(9*4/3 π*r^3)
= (125-12π)/12π
≈ 2.32
Huh? The planetary radius is equal to the separation? Do the planets touch, or is the separation between planetary surfaces?
If so, then a cube with side 5r contains about 9 planets of radius r. So, the void ratio would be
((5r)^3 - 9*4/3 π*r^3)/(9*4/3 π*r^3)
= (125-12π)/12π
≈ 2.32
0 answers