Gravity problem: determining the period of a satellite in Earth orbit

[BiggestAfrica]

Homework Statement

Given: G = 6.67259 × 10−11 N m2/kg2
The acceleration of gravity on the surface of a planet of radius R = 4430 km is 6.23 m/s^2. What is the period T of a satellite in circular orbit h = 15372.1 km above the surface? Answer in units of s.

Relevant Equations

(Not assigned any specific equations, but these are the ones I think relate to the problem)
a = (v^2)/r
F = G(m1)(m2)/r^2
T = 2(pi)(r)/v

Hi!

This is a problem from my physics 1 high school course. I've tried using the first and third equations to determine period (answer of 8326.9544s.), however that was incorrect and I never even touched G. I'm not sure where to go from here at all. Any help is appreciated!

 

[PeroK]

How did you calculate the velocity?

 

[BvU]

Hello BA, !

a = (v^2)/r
F = G(m1)(m2)/r^2
T = 2(pi)(r)/v

I've tried using the first and third equations

I take it you can find $r$, but how do you determine $v$ or $a$ ?
Can you show your work ?

 

[BiggestAfrica]

To find velocity, I rewrote a = v^2/r
v = sqrt(6.23*4430000) = 5253.4655m/s
a= 6.23m/s^2

 

[PeroK]

To find velocity, I rewrote a = v^2/r
v = sqrt(6.23*4430000) = 5253.4655m/s
a= 6.23m/s^2

That looks like a satellite orbiting at the surface!

 

[BiggestAfrica]

Oops! I had tried using both radii and the different between them for the final calculation, but not the velocity. It is my last try on this HW, so just to be sure, I use the height for the calculations, or the difference between the height and radius of the planet?

 

[PeroK]

Oops! I had tried using both radii and the different between them for the final calculation, but not the velocity. It is my last try on this HW, so just to be sure, I use the height for the calculations, or the difference between the height and radius of the planet?

Neither. The radius is the radius of the orbit. Which is the distance from the centre of the planet.

But, you are also missing a calculation of the gravitational acceleration at the orbit. That's not the same as at the surface.

 

[Janus]

The orbit is nearly 4.5 times further from the center of the planet than the surface. What happens to the value of the acceleration due to gravity as you move away from the planet?

 

[BiggestAfrica]

Sorry for the delayed response, but I took your suggestions in mind and here's what I tried to no success.

Found the mass of the planet with the equation a = Gm/r^2
6.23 = G(m)/(4430000^2)
m = 1.8323e24

Used the mass to find acceleration of gravity at h = 15372.1km
a = G(1.8323e24)/(15372100^2)
a = 0.5174

Then I found the period with the equation T = 2(pi)(r)/sqrt(a*r)
T = 2(pi)(15372100)/sqrt(.5174*15372100)
T = 34247.7402 seconds

I'm not sure where I went wrong?

 

[PeroK]

Used the mass to find acceleration of gravity at h = 15372.1km
a = G(1.8323e24)/(15372100^2)
a = 0.5174

I'm not sure where I went wrong?

In the formula for gravitational acceleration: $g = \frac{GM}{r^2}$, $r$ is the distance from the centre of the planet. The height above the surface that you are using is not the distance from the centre: $h \ne r$.

$r = h + R$

 

[BiggestAfrica]

In the formula for gravitational acceleration: $g = \frac{GM}{r^2}$, $r$ is the distance from the centre of the planet. The height above the surface that you are using is not the distance from the centre: $h \ne r$.

$r = h + R$

Thank you that was exactly it!