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Transcript
```Proportionality between the
In circular motion with a constant
centripetal force.
Applying Newton’s 2nd Law
• 𝐹 = 𝑚𝑎
• In this case the only force on the body is the
force of gravity and the acceleration of the
body is centripetal.
• 𝐹𝐺 = 𝐹𝐶
A centripetal force is required for circular
motion. This is given by the equation:
v2
Fc  m1 (
)
r
The centripetal force is supplied by the
force of gravity between the two bodies
which is given by the equation:
Gm1m2
FG 
r2
Set the two forces equal to one another.
FC  FG
2
Gm1m2
v
m1

2
r
r
Gm1m2
v2
m1

2
r
r
m1 = mass of the satellite
m2 = mass of object being orbited
V = velocity of the satellite
G = universal gravitational constant
r = radius of the circular motion
Mass of the satellite cancels out. The speed
of the satellite depends on radius r and mass
m of the object being orbited.
Orbital Velocity
Solving for v
Gm
v
r
Where:
v = orbital velocity (m/s)
G = universal gravitational constant = 6.67 x 10-9
m = mass of the object being orbited (kg)
r = radius of circular motion (m)
Period of Satellite
• We know that
x
x
v  or t 
t
v
For a circular motion, the distance x travelled by the
satellite in one revolution is the circumference of the
circle. That would be
x  2r
Solving for t
x
t
v
or
2r
t
v
But we also know that
This gives us
t 
Gm
v
r
2r
Gm
r
Squaring both sides
2 2
4

r
2
t 
Gm
r
We end up with orbital period t of a
satellite. This is the time it takes for a
satellite to complete one orbit.
t  2
3
r
Gm
Again, m here is the mass of the object being orbited
NOT the mass of the satellite.
Quick Check # 1:
• What is orbital velocity of the earth around
the sun? The sun has a mass of 1.99 x
1030kg, the mean distance from the earth
to the sun is 1.50 x 1011 m.
Quick Check #1 Solution
v
2
N.m
30
6.67 x10 11
(1.99
x
10
) kg
2
Gm
kg

r
1.50 x 1011 m
2
m
8
4 m
v  8.85 x10 2  2.97 x 10
s
s
Quick Check #2
• A satellite is in a low earth orbit, some 250 km
above the earth's surface. rearth is 6.37 x 106 m and
mearth = 5.98 x 1024 kg. Find the period of the
satellite in minutes.
Quick Check #2 Solution
r is the distance from the center of the earth to the
satellite; thus
r = rearth + 250 km = 6.37 x 106 m + 250 x 103 m
T  2
3
r
Gm

 2
6.67 x 10
 1 min 
T  5357 s 

60
s


6.62 x 106 m

3

m m2
24
5.98
x
10
kg
2
2
s kg
11 kg

89.3 min

```
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