# Homework Help: Centripetal forces and satellites

1. Aug 22, 2011

### supmiller

Hi im kind of new to the forum nice to be here! here is my question.

1. The problem statement, all variables and given/known data

A satellite is kept in a circular orbit 300km above the surface of the earth by the force of gravity. At this altitude the acceleration due to gravity is 8.9 m/s. The radius of the earth is 6.4*10^6m

a) calculate the period of the satellite

b) calculate the speed of the satellite

2. Relevant equations

Ac=(4(Pi^2)(r))/(T^2)

Where ac is the acceleration
T is the period
r is the radius

V=(2(pi)r)/(T)

3. The attempt at a solution

a)

Using the first equation listed

T= (2(pi))(r/ac)^(1/2)
r=(6.4*10^6)+(300km)(1000m/1km)
ac=8.9m/s

T=5451.6 seconds

b) Using our known T,
V=(2Pi)(6700000)/(5451.6)
V=7722 m/s

Thanks in advance, I feel I may not have the concept down well enough and this may not be the way to do it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 22, 2011

### PhanthomJay

That looks right to me. Welcome to PF as a poster!!

3. Aug 23, 2011

### c03rcion

** r = radius
r = RE + h = (6.4*10^6m + 300,000m) = 6.7x10^6

**C = Circumference
C=2(PI)r

T=period
T=C/v

**for circular motion there are 2 components of acceleration, radial acceleration and tangential acceleration. the radial (centripetal acceleration) is shown:

Ar=(mv2)/r

**because gravity from the Earth is causing the satellite to undergo centripetal acceleration:

mg = (mv2)/r

**the masses of the satellite cancel to show the speed of the satellite:

(rg)1/2=v

v = 7722 m/s

** the period
T = 2(PI)r/v = 5448s

Based on what I got, I would say you did a good job!

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