
I have tried to calculate the moon's gravity at
100,000 km distance from its surface by using the
inverse square law, but I get confusing results.
Please help me. 
Question Date: 20130626   Answer 1:
You can read your answer to this question on
the following link:
moon&grav.png
  Answer 2:
The gravitational force between two bodies of
mass m1 and m2 should be F = G*m1*m2 /
r^{2}.
The mass of the moon = 7.346
*10^{22} kg, G = 6.67 *10^{11}
m^{3}/kgs^{2},and you stated that
r = 100,000 km (the radius of the moon is only
~2,000 km so we won't worry that we are measuring
r from the surface not the center.)
From this, we can say the force on a 1 kg
object would be 0.00049 N, or 0.000011 lbs. In
other words, this would be 20,000 x less than the
gravity we feel on the surface of the earth.
This should make sense, the mass of the moon is
~80 times smaller than that of the earth. The
radius of the earth is ~6,500 km, so your r is 16
times larger.
Thus from before for
F=Gm1*m2/r^{2}, if we decrease m2 by a
factor of 80, and increase r by a factor of 16,
the new force should be (1/80) / (16)^{2}
= 1/20,000 what we feel on earth.
  Answer 3:
The acceleration due to gravity at distance
R from the center of a spherical body is
GM/R^{2} where G is the universal
gravitational constant, M is mass of moon or
planet, and R is the distance from the center of
the body in a radial direction.
At the surface of the body it would be the
body's radius,
so for Earth it is g=G*Mearth/Rearth^{2},
but if one was 1000 miles above the Earth surface,
it would be Rearth + 1000 miles , etc.
  Answer 4:
Let’s take a look at Newton’s law of universal
gravitation to help answer your question.
Newton’s law states that the force (F) enacted
on two bodies interacting with each other due to
gravity is the product of the gravitational
constant (G), the masses of the two bodies (m1 and
m2) and the inverse of the distance squared
between their center of mass (1/r^{2}).
The complete equation is:
F = G*m1*m2/r^{2}.
If we take out the mass of the variable body (i.e.
keep the mass of the moon in the equation), the
equation reduces to the acceleration (a) due to
the gravitational force of the remaining body.
You can verify that the acceleration at the
surface of the Earth should be 9.8 m/s^{2}
(one G) by using these numbers (from
hypertextbook.com; make sure units correctly
cancel out!):
G = 6.7X10^{11} m^{3}/kg/s
m_earth = 6X10^{24} kg
r = (diameter_earth)/2 = (12800 km) / 2 = 6400 km
= 6.4X10^{6} m
Thus, a = G*m_earth/r^{2} = 9.81
m/s^{2} (correct!)
So, now we can calculate the acceleration of an
object 100,000 km from the surface of the moon.
We need to make sure to add the distance from the
center of the moon to this 100,000 km (numbers
again from hypertextbook.com):
m_moon = 7.5X10^{22} kg
r = 100,000 km + (diameter_moon)/2 = 100,000 km +
(3480 km)/2 = 1.02X10^{8} m
Thus, a = G*m_moon/r^{2} =
4.8X10^{4} m/s^{2}
The results should be expected; we are looking at
the gravitational effects of something that is
very, very far away from the moon (about 1/4 the
distance from the Earth to the moon). But while
the numbers seem very small, they are not
negligible. The small shifts in gravitational
force are indeed enough to cause changes on Earth
such as the rise and fall of tides due to the moon
(and Sun).
  Answer 5:
The inverse square law is a very interesting one
in that it shows up everywhere in physics. As
applied to gravity the law can be written as such:
F = G{m1*m2}/r^{2}.
F is the force in newtons, G is the gravitational
constant (6.674×10^{11} N m^{2}
kg^{2})
m1 and m2 are the masses of the two things being
attracted, in your case, the moon with a mass of
7.347 x 10^{22} kg, and some other mass
(in kg) and the distance between the masses, in
meters.
The reason you need two masses is because gravity
can only be felt by its effect
on matter (like all fundamental forces) and since
gravity only works on mass, we can only measure it
based on its effect on
something with mass. Two massive objects will
attract each other with equal force. So, that is
to say, when you are falling the earth's
gravity is exerting a certain force on you, but
you are exerting the exact same force on the
earth. The difference is that the earth is so
large that it really doesn't move very much at
all. (This is because F = MA, force equals mass
times acceleration. If the force is the same
then your acceleration will be much higher than
the earth's because your mass is so much lower
than the earth's mass).
So the problem you may be having could be the
value you use for G, could be the second mass that
you use, or it might be the units you're using for
all the values. I hope that helps.
Click Here to return to the search form.





Copyright © 2017 The Regents of the University of California,
All Rights Reserved.
UCSB Terms of Use


