Period of satellite in circular orbit

[bmoore509]

Homework Statement

Given: G = 6.67259 × 10−11 Nm2/kg2
The acceleration of gravity on the surface
of a planet of radius R = 4190 km is 6.5 m/s2.
What is the period T of a satellite in circu-
lar orbit h = 14120.3 km above the surface?
Answer in units of s.

Homework Equations

a=G(m/r^2)
T = 2πR/V
V = sqrt(GM/R)
T = 2πR/sqrt(GM/R)

The Attempt at a Solution

My formulas might be incorrect but I really don't think they are. Yet I can't seem to get the right answer.

6.5 m/s² = (6.67259 × 10⁻¹¹ Nm²/kg²) (m/((4190000m+14120300m)^2))
m=3.265952291x10²⁵

T = 2piR/sqrt(GM/R)

= 2*pi*(4190000+14120300)/ sqrt((6.67259 × 10⁻¹¹ *3.265952291x10²⁵)/(4190000m+14120300m)

=1054559091seconds

But that answer isn't right. Any help please?

 

[Filip Larsen]

The equations look correct, so I guess you made a calculation error in the last expression.

Note, that the period usually is written with the two R's combined into one, like
$$T = 2\pi\sqrt{\frac{R^3}{GM}}$$

 

[ehild]

6.5 m/s² = (6.67259 × 10⁻¹¹ Nm²/kg²) (m/((4190000m+14120300m)^2))

6.5 m/s^2 is the acceleration at the surface of the planet, r=4190 km.

ehild

 

[bmoore509]

I redid my work using the formula as you wrote it.

T = 2pi (sqrt((R^3)/GM))

R³= (4190000+14120300)³ = 6.138840926x10²¹

G*M = (6.6725x10^-11)*(3.265952291x10²⁵)=2.17923606x10¹⁵

sqrt ( R³/(G*M) = sqrt (6.138840926x10²¹/2.17923606x10¹⁵) = sqrt (281696.23) = 1678.382921

T=2pi(sqrt(r^3/GM)
= 2*pi*1678.382921 = 10545.59091

Which isn't the right answer. Can you tell me where I'm going wrong?

 

[bmoore509]

Oh! Thank you! Let me rework this then.

 

[bmoore509]

T = 2pi (sqrt((R^3)/GM))

R3= (4190000+14120300)³ = 6.138840926x1021

G*M = (6.67259x10^-11)*(1.710200237x10²⁴)=1.1411465x10¹⁴

sqrt ( R3/(G*M) = sqrt (6.138840926x10²¹/1.1411465x10¹⁴) = sqrt (53795379.7) = 7334.533366

T=2pi(sqrt(r^3/GM)
= 2*pi*7334.533366 = 46084.23228


Does that look correct?

 

[ehild]

Yes, if you mean seconds. But use only 6 significant digits, and write out the units.

ehild

 

[Filip Larsen]

6.5 m/s^2 is the acceleration at the surface of the planet, r=4190 km.

Ah, I missed that the height wrongly was "included" in the calculation of planet mass. My apologies to the original poster.