Balanced forces<span> act on the same object and </span>Action-Reaction forces<span> act on different objects.</span>
Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
Answer:
16294 rad/s
Explanation:
Given that
M(ns) = 2M(s), where
M(s) = 1.99*10^30 kg, so that
M(ns) = 3.98*10^30 kg
Again, R(ns) = 10 km
Using the law of gravitation, the force between the Neutron star and the sun is..
F = G.M(ns).M(s) / R²(ns), where
G = 6.67*10^-11, gravitational constant
Again, centripetal force of the neutron star is given as
F = M(ns).v² / R(ns)
Recall that v = wR(ns), so that
F = M(s).w².R(ns)
For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence
F = F
G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)
Making W subject of formula, we have
w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]
w = √[{G.M(ns)} / {R³(ns)}]
w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]
w = √[2.655*10^20 / 1*10^12]
w = √(2.655*10^8)
w = 16294 rad/s
the earth moves throughout the year such as rotate around the sun, so yes the it does move and it sits roughly at 93.048 million miles away from the sun. I hope this helps you out! :)
plastics, Styrofoam hOPE THIS HELPS
