(1) On earth, two parts of a space probe weigh 16500 N and 4800 N. These parts a
ID: 1977174 • Letter: #
Question
(1) On earth, two parts of a space probe weigh 16500 N and 4800 N. These parts are separated by a center-to-center distance of 11 m and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects. ___________N(2) A bowling ball (mass = 7.2 kg, radius = 0.13 m) and a billiard ball (mass = 0.37 kg, radius = 0.028 m) may each be treated as uniform spheres. What is the magnitude of the maximum gravitational force that each can exert on the other?___________N
(3) A space traveler weighs 570 N on earth. What will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth? ___________N
Explanation / Answer
1) we just use the general equation for gravitational force.
First we have to convert the weight of the probes to masses.
16500N/9.8m/s2=1683.67 Kg
4800N/9.8m/s2=489.80 Kg
FG=Gm1m2/r2=6.673*10-11(1683.67)(489.80)/112
FG=4.55*10-7 N
2)The maximum gravitational force will occur when they are as close as possible. This distance ends up being the sum of their radii. So we plug the information into the gravity equation.
FG=Gm1m2/r2=6.673*10-11(7.2)(.37)/(.13+.028)2
FG=7.121*10-9 N
3)Alright so we know:
570=Gm1m2/r2=6.673*10-11Mem/re2
Now if we put in 2Me and 3re we get
F=6.673*10-11(2Me)m/(3re)2
F=6.673*10-11Mem/re2[2/9]
Notice the two underlined parts are equal.We know this is equal to 570 from the first equation.
F=570[2/9]
F=126.67N
Hope that helps
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