# Gravity And Circular Motion

by Karel Celustka, February 1999

Here's the relevant quantities:

• r distance between centers
• m1 mass of central object
• m2 mass of satellite
• Fg force due to gravity
• Fc centripetal force
• G universal gravitational constant = 6.67(10-11)Nm2/kg2
• v orbital velocity
• T orbital period
• ac centripetal acceleration
These quantities are explained on other pages.

## New Quantities

Sorry! There are no new quantities here.

## Formulae

Here are the relevant formulae that we have so far:

`ac = V2/r = 4(p)2r/T2`
`(T1/T2)2 = (r1/r2)3`
`Fg = Gm1m2/r2`
1. Fc = m2v2/r
2. v/r = 2pi/Tand now for the only new thing here:
3. For the orbital condition to be satisfied, Fg = Fc

4. Simple substitution then gets us to the most common variant, Gm1m2/r2=m2w2r.
This simplifies to v2r = Gm1
Note that the satellite's mass drops out here. Substituting 2p/T for v/r leads to
Kepler's observation that T2/r3 is constant for all planets in our solar system.

## General Problem Solving Strategy:

2. Go through the problem and figure out what is given or implied

3. Make a list, and identify the quantities you know.
4. Set up the force of gravity equation or the centripetal acceleration equation.
5. Solve it for what you're looking for. Replace known values with their numerical equivalents. For this type of problem, this is all you should ever need to do.
6. WARNING: Watch for whether the radius between objects is from center to center, not edge to edge.

## Example problem 1

What is the force of gravity between the two 4.2 Kg bowling balls whose centers are 1.2 m
distant?

• m1 = 4.2 kg
• m2 = 4.2 kg
• r = 1.2 m
First, you need to realize that Fg = Gm1m2/r2. Then, you basically stuff the knowns into the
equation. So, Fg =(6.67*10-11 N*m2/kg2)(4.2 kg)(4.2 kg)((1.2)-2). The answer you should get
is 8.174*10^-10 N.

## Example problem 2

What is the force of gravity between the earth and the moon? What acceleration does the moon
undergo? (Hint: use F=ma) What is its period of motion? (Use a centripetal acceleration equation)

First, you need to realize again, that Fg = Gm1m2/r2 and that m1 =29.11*10^24 kg (earth's mass)
and m2 =7.2*10^22 kg (moon's approx. mass). Next, we need to figure out what the distance is
from the center of the earth to the center of the moon. Looking this up, we find the crust-to-crust
distance is 3.85*10^8 m, the radius of the moon is 1.74*10^6 m, and the radius of the earth is
6.37*10^6 m. Now we add all this up and get that r=3.93*10^8 m. Now we can just put in the
numbers: Fg =(6.67*10-11 N*m2/kg2)(29.11*10^24 kg)(7.2*10^22 kg)((3.93*10^8 m)^-2).
The answer you should get is 2.03*10^20 N.

For the next part, take the answer from the first part 2.03*10^20 N and divide it by the moon's
mass of 7.2*10^22 kg. You should get [2.03*10^20 N=(7.2*10^22 kg)a] that a=2.8*10^-3
m/s/s.

Next, we want to use the equation ac = 4(pi)2r/T2, and we know a=2.8*10^-3 m/s/s. And we
know that r=1.74*10^6 m. Now, solve this equation. 2.8*10*-3 m/s/s=4(pi)2(1.74*10^6
m)/(T^2). You should get that T=2.3*10^6 s or 27 days.

## Example problem 3

What is the force of gravity on a 10 kg object twice earth's radius above the earth?

First, find the earth's radius. It is 4.7*10^6 m, and since the 10 kg object is twice earth's radius
above the earth, r=1.41*10^7 m. We also need the earth's mass, it is 29.11*10^24 kg. Now,
we just stuff all this into the force of gravity equation. Fg =(6.67*10-11 N*m2/kg2)(29.11*10^24
kg)(10 kg)((4.22*10^7 m)^-2). You should get that the force of gravity is 10.9 N.

## Example problem 4

What distance from the center of the moon is the attraction between a 500 kg object and the
moon itself equal
to 2.5 N?

We know that Fg =2.5 N, the m=500 kg, the moon's m=7.2*10^22 kg. Now, stuff all knowns
into the force of gravity equation and solve: 2.5 N=(6.67*10-11 N*m2/kg2)(7.2*10^22 kg)(500
kg)((r)^-2). You then should find that r=3.1*10^7 m.

## Sample Problems

```
The answers to each problem follow it in parentheses.  They also link to a solution to the problem.```
```Try the problem, check your answer, and go to the solution if you do notunderstand.

```

## 1.

```What is the centripetal acceleration of a car going 21 m/s around a corner with a radius of 45 m?

(9.8 m/s/s )

```

## 2.

```What must be the radius of curvature for a curve in a road if cars going 27 m/s never exceed a lateral acceleration of 4.2 m/s/s?

(173.6 m)

```

## 3.

```My econo-box pulls .72 g's (multiply .72 by 9.8) of lateral acceleration going around a corner witha radius of 35 m. What is its speed?

(15.7 m/s)

```

## 4.

```A centrifuge has a tangential velocity of 34 m/s and a radius of 12 cm. What is the centripetal acceleration at its edge?

(9633 m/s/s)

```

## 5.

```A certain car can pull .92 g's of lateral acceleration. What radius turn can it do at 40 m/s?

(177.5 m)

```

## 6.

```Your 1963 microbus can do about 4.5 m/s/s of centripetal acceleration. How fast dare you go around a corner with a radius of 60 m?

(8.8 x 107m)

```

## 7.

```What centripetal force is needed to make a 5 kg hammer swing 3.4 m/s in an arc with radius 1.75 m?

(33 N)

```

## 8.

```What centripetal force is needed to make a 65 kg rider go 15 m/s on the edge of a ride with radius of 4.5 m?

(3250 N)

```

## 9.

```A 320 kg space probe has jets which can exert a centripetal force of 120 N. What is the sharpest radius of a turn it can make if

it is going 520 m/s?

(7.2*10^5 m/s)

```

## 10.

```How fast can your 800 kg car go around a corner with a radius of 13 m when the available centripetal force is 6500 N?

(10.3 m/s)

```

## 11.

```What is the force of gravity between the two 4.2 kg bowling balls whose centers are 1.2 m distant?

(8.17*10^-10 N)

```

## 12.

```What is the force of gravity between the earth and the moon?

(2.03*10^20 N)

```

## 13.

```What is the force of gravity on a 10 kg object twice earth's radius above the earth?

(10.9 N)

```

```
```

## 1.

```The centripetal acceleration=v^2/r. So, (21^2)/45=9.8 m/s/s.

(Go To Problem )

```

## 2.

```We know v=27 m/s, r=? and ac=4.2 m/s/s. Set up the equation and solve: 4.2=(27^2)/r. So r=173.6 m.

(Goto Problem)

```

## 3.

```We know v=?, r=35 m and ac=(.72*9.8) m/s/s. Set up the equation and solve: 7.056=(v^2)/35. So, v=15.7 m/s.

(Goto Problem)

```

## 4.

```We know v=34 m/s, r=.12 m and ac=? . Set up the equation and solve: ac=(34^2)/.12. So, ac=9633 m/s.

(Goto Problem)

```

## 5.

```We know v=40 m/s, r=? and ac=(.92*9.8) m/s/s. Set up the equation and solve: 9.016=(40^2)/r. So, r=177.5 m.

(Goto Problem)

```

## 6.

```We know v=? , r=60 m and ac=4.5 m/s/s. Set up the equation and solve: 4.5=(v^2)/60. So, v=16.4 m/s.

(Goto Problem)

```

## 7.

```We know v=3.4 m/s , r=1.75 m, m=5 kg and Fc=? . Set up the equation and solve: Fc=(5)(3.4^2)/1.75. So, Fc=33.03 N.

(Go To Problem )

```

## 8.

```We know v=15 m/s , r=4.5 m, m=65 kg and Fc=? . Set up the equation and solve: Fc=(65)(15^2)/4.5. So, Fc=3250 N.

(Goto Problem)

```

## 9.

```We know v=520 m/s , r=? , m=320 kg and Fc=120 N . Set up the equation and solve: 120=(320)(520^2)/r. So, r=7.2*10^5 m.

(Goto Problem)

```

## 10.

```We know v=? , r=13 m , m=800 kg and Fc=6500 N . Set up the equation and solve: 6500=(800)(v^2)/13. So, v=10.3 m/s.

(Goto Problem)

```

## 11.

```Go and see Example #1

(Goto Problem)

```

## 12.

```Go and see Example #2

(Goto Problem)

```

## 13.

```Go and see Example #3

(Goto Problem)