Finding the astronaut's weight on a planet's surface

[Vanessa Avila]

Homework Statement

A landing craft with mass 1.22×10^4 kg is in a circular orbit a distance 5.50×10^5 m above the surface of a planet. The period of the orbit is 5100 s . The astronauts in the lander measure the diameter of the planet to be 9.50×10^6 m . The lander sets down at the north pole of the planet.
What is the weight of an astronaut of mass 84.1 kg as he steps out onto the planet's surface?

Homework Equations

circular motion speed: v = √GM/r
g: GM/r²
T = 2πr/v

The Attempt at a Solution

I tried to solve for the speed by using the period.
5100 = 2π(5.50*10^5)/v
v = 677.6 m/s
and using the circular motion speed equation, I tried to solve for M, where the radius i used is the radius of the planet:
677.6 = √6.67*10^-11*M/4.75*10^6
M = 3.27*10^22

and then i used GM/r^2 to solve for G.

6.67*10^-11*3.27*10^22/(4.75*10^6) and i got a really small g. I think I'm missing something. Especially because I haven't used the space satellite's mass.

 

[TSny]

A landing craft ... in a circular orbit a distance 5.50×10^5 m above the surface of a planet.
T = 2πr/v
5100 = 2π(5.50*10^5)/v

In the equation T = 2πr/v, what does r represent? What does the distance 5.50×10^5 m represent?

 

[Vanessa Avila]

In the equation T = 2πr/v, what does r represent? What does the distance 5.50×10^5 m represent?

That distance represents how far the landcraft is hovering above the planet

 

[TSny]

That distance represents how far the landcraft is hovering above the planet

That's what the number 5.50 x 10⁵ m represents. But the r in 2πr/v is a different distance.

 

[Vanessa Avila]

That's what the number 5.50 x 10⁵ m represents. But the r in 2πr/v is a different distance.

so the r should be the radius of the planet instead of the distance the landing craft is from the planet?

 

[TSny]

so the r should be the radius of the planet instead of the distance the landing craft is from the planet?

No. Think about where the formula T = 2πr/v comes from. The craft moves at constant speed while in orbit. For something moving at constant speed, distance = speed x time. Or, time = distance/speed. Apply this to one orbit of the craft so that the time is equal to the period T:

T = distance/v. Here, the distance is how far the craft travels in one circular orbit.

 

[Vanessa Avila]

No. Think about where the formula T = 2πr/v comes from. The craft moves at constant speed while in orbit. For something moving at constant speed, distance = speed x time. Or, time = distance/speed. Apply this to one orbit of the craft so that the time is equal to the period T:

T = distance/v. Here, the distance is how far the craft travels in one circular orbit.

so would the r be the planet's radius + landing craft's distance?

 

[TSny]

so would the r be the planet's radius + landing craft's distance?

Yes. r is the radius of the craft's orbit.

 

[Vanessa Avila]

can i use > 5100s = 2π(r1+r2)/√GmM/(r1+r2) to solve for M ?

Yes. r is the radius of the craft's orbit.

 

[TSny]

can i use > 5100s = 2π(r1+r2)/√GmM/(r1+r2) to solve for M ?

Yes, except why does m appear on the right side?

Nit picky note: To avoid confusion with the square root symbol √, use parentheses to show what belongs inside the square root. If I write √ab, then it is not clear if both the a and the b are inside the square root. Most people would interpret √ab to mean that only the "a" is inside the square root. If you want to indicate that both a and b are inside, then you should write √(ab).