What is the Center of Mass of the Earth + Satellite System?

AI Thread Summary
The discussion focuses on calculating the center of mass of the Earth and a satellite system, with the satellite's mass being 2970 kg and its orbit at 42400 km above Earth's surface. The correct formula for the center of mass is highlighted, emphasizing that all distances should be measured from the Earth's center. A common mistake noted is the incorrect addition of the Earth's radius to the calculated center of mass, which should not be necessary since the calculation already accounts for the Earth's center. Participants are reminded to consider the relative masses involved, as the satellite's mass is negligible compared to that of the Earth, leading to minimal displacement of the center of mass. Understanding these principles is crucial for accurately solving the problem.
smpolisetti
Messages
10
Reaction score
0
This problem is driving me crazy!

Homework Statement


How far from the center of the Earth (in km) is the center of mass of the Earth + satellite system if the satellite has a mass of 2970 kg and is in orbit 42400 km above the surface of the Earth. The mass of the Earth is 5.98*10^24 and the radius of the Earth is 6370

So:
Mass of earth: 5.98*10^24 kg
Mass of satellite: 2960 kg
Distance of satellite from surface of the earth: 42400
Radius of earth: 6370

Homework Equations


Center of mass = (x1m1 + x2m2)/ (m1+m2)
m= mass
x= distance



The Attempt at a Solution


Here was my attempt

I set up center of mass = [(0km (5.98*10^24)) + (2970kg * 42400km) ] / ((5.98*10^24)+2970)

From that I got 2.106*10-17 km

Which I then added to the radius of the Earth to get 6370 km

But that's wrong, what was my mistake?
 
Physics news on Phys.org
smpolisetti said:
I set up center of mass = [(0km (5.98*10^24)) + (2970kg * 42400km) ] / ((5.98*10^24)+2970)

All distances in the problem are being measured relative to the centre of the Earth. So, this distance in red should be the distance between the satellite and the centre of the Earth.

smpolisetti said:
From that I got 2.106*10-17 km

Which I then added to the radius of the Earth to get 6370 km

Why would you do this? :confused: All distances in this problem are referenced to the centre of the Earth. That means that the answer you get from your equation will be the position of the centre of mass relative to the centre of the Earth, which is exactly what you want. No corrections are necessary.

Also, please THINK about your answers and whether they make sense. Use your intuition. If the satellite weren't there, where would the centre of mass of the Earth be? It would be at the centre of the Earth, right? So, is adding a satellite, an object whose mass is negligible compared to the planet, really going to shift the centre of mass very far away from the centre of the planet?
 
Thanks a ton!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top