Physics: Gravitational Force Circular Motion

In summary: F=(GMm)/r^2 to get a=(GM)/r^2 = (4 pi^2 r^2)/T^2, then you can isolate for r to get r=the cube root of [(GMT^2)/4pi^2)]
  • #1
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Homework Statement



A.) What is the acceleration due to gravity at an altitude of 1500 km above the Earth's surface. Neglect the effect of the Earth's rotation. The Earth's radius is 6378100 m and its mass is 5.98 * 10[tex]^{24}[/tex]kg.

B.) The moon has a mass of 7.35 * 10[tex]^{22}[/tex] kg and is located 3.84 * 10[tex]^{8}[/tex] m from the Earth (center to center). At some point between the Earth and the Moon, the gravitational force due to the Earth on any object cancels the gravitational force due to the Moon. Where is that position, relative to the center of the Earth?

C.) The asteroid belt consists of so many particles, we haven't counted them all yet. However, more than half of the mass of the belt is contained in the four largest asteroids. The largest - Ceres - takes 4.6 years on average to circle the Sun. What is the orbital radius of Ceres in meters.

1 year = 365 days, 1 day=24 hours, 1 hour=3600 seconds

Homework Equations



F=(GMm)/r[tex]^{2}[/tex]
F=ma
a=v[tex]^{2}[/tex]/r
v=(2[tex]\pi[/tex]r)/T
G=6.67 * 10[tex]^{-11}[/tex]
Sun's Mass = 1.99 * 10[tex]^{30}[/tex]

The Attempt at a Solution



A.) (GMm)/r[tex]^{2}[/tex]=(mv[tex]^{2}[/tex])/r
the m's cancel out and you are left with a (which was equal to v[tex]^{2}[/tex]/r) = (Gm)/r[tex]^{2}[/tex]
so a = [tex]\underline{(6.67*10[tex]^{-11}[/tex])(5.98 * 10[tex]^{24}[/tex] }[/tex] (1500+6378100)

B.) I really don't understand what this question is asking. I think that it means that there is a place between the moon and the Earth that an object has no gravity. Would that mean that
(GM[tex]_{earth}[/tex]m)/r[tex]^{2}[/tex][tex]_{x}[/tex] = (GM[tex]_{moon}[/tex]m)/r[tex]^{2}[/tex][tex]_{x}[/tex] then everything just cancels out.

C.) (GMm)/r[tex]^{2}[/tex]=(mv[tex]^{2}[/tex])/r the m's cancel out. Then you replace v^2 with [tex]\underline{4[tex]\pi^{2}[/tex]r}[/tex]/T[tex]^{2}[/tex]. And then you move the r to the other side. So

r=[tex]\sqrt[3]{GMT^{2}}[/tex]/ [tex]\sqrt[3]{4\pi^{2}}[/tex]


Thanks for any help.
 
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  • #2
Sorry, for my attempt

A.) a=(GM)/r^2

C.) r=the cube root of [(GMT^2)/4pi^2)]
 
  • #3
for C.) v=(2 pi r)/T so v^2 = (4 pi^2 r^2)/T^2
 

Related to Physics: Gravitational Force Circular Motion

1. What is the formula for gravitational force?

The formula for gravitational force is F = G(m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

2. How does the force of gravity affect circular motion?

The force of gravity acts as a centripetal force in circular motion, constantly pulling an object towards the center of the circular path. This allows the object to maintain its speed and direction in the circular motion.

3. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It is always directed towards the center of the circle and is equal to the product of the object's mass, its speed, and the radius of the circular path.

4. How is gravitational force related to the mass and distance between two objects?

Gravitational force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them. This means that as the masses increase, the force of gravity increases, and as the distance between them increases, the force decreases.

5. What is the difference between gravitational force and gravitational acceleration?

Gravitational force is the attraction between two objects due to their masses, while gravitational acceleration is the acceleration experienced by an object due to the force of gravity. Gravitational acceleration is dependent on the mass of the larger object and the distance between the two objects.

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