At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
The orbit is a circle whose radius is 3 times the radius of the surface
(both measured from the center of the moon). So the acceleration due
to gravity at the orbital altitude is
1/3² = 1/9 = 11.1% of its value on the surface.
(both measured from the center of the moon). So the acceleration due
to gravity at the orbital altitude is
1/3² = 1/9 = 11.1% of its value on the surface.
From Newton's Law of Universal Gravitation:
F = GMm / r^2. ( When at the surface).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Moon
m = Mass of Satellite
r = distance apart, between centers = in this case it is the distance from Moon to the Satellite.
Recall: F = mg.
mg = GMm / r^2
g = GM / r^2.........................(i). When at surface.
Note when the satellite is at a distance 2 times the radius of the moon.
Therefore, the distance between centers = 2r + r = 3r.
Note, when need to add radius of the moon, because we are measuring from center of the satellite to center of the moon.
From (i)
g = GM / (3r)^2. The distance r is replaced with 3r
g = GM / 9r^2 = (1/9) * GM / r^2
Therefore gravity on the satellite is (1/9) times that on the Moon.
F = GMm / r^2. ( When at the surface).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Moon
m = Mass of Satellite
r = distance apart, between centers = in this case it is the distance from Moon to the Satellite.
Recall: F = mg.
mg = GMm / r^2
g = GM / r^2.........................(i). When at surface.
Note when the satellite is at a distance 2 times the radius of the moon.
Therefore, the distance between centers = 2r + r = 3r.
Note, when need to add radius of the moon, because we are measuring from center of the satellite to center of the moon.
From (i)
g = GM / (3r)^2. The distance r is replaced with 3r
g = GM / 9r^2 = (1/9) * GM / r^2
Therefore gravity on the satellite is (1/9) times that on the Moon.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.