Solved: Synchronous Orbit Altitude of Mercury in km

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Homework Help Overview

The problem involves determining the altitude of a synchronous orbit for a satellite over Mercury, requiring the application of gravitational equations and unit consistency.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate units for the rotational period and the implications of using different unit systems. There is a focus on the relationship between mass, gravitational constant, and the orbital radius.

Discussion Status

Participants are actively engaging with the problem, clarifying the need for consistent units and exploring the correct period to use for Mercury. Some guidance has been provided regarding unit conversions and dimensional analysis, but no consensus has been reached on the final approach.

Contextual Notes

The original poster expresses confusion about the appropriate units for the rotational period and how to achieve the desired units for the final result. There is an emphasis on ensuring that the calculations align with the gravitational constant used.

shade585
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[SOLVED] synchronous orbit

Homework Statement



A orbiting satellite stays over a certain spot on the equator of (rotating) Mercury. What is the altitude of the orbit (called a "synchronous orbit")?

The answer needs to be in km.

Homework Equations



I know the synchronous orbit is where the satellite has a period equal to the average roational period of the planet.

T^2 = ((4pi^2)/(GM))r^3

T would be the rotational period of Mercury.

G is the gravitational constant.

M would be mercury's mass.

I would be solving for r.

The Attempt at a Solution



So far i have been plugging in numbers but nothing has been working. Should T be in years, days, hours, or seconds?
 
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The units for the rotation period must agree with the units you use for G.
 
Equation

The last I checked. the mass term is not in the equation. I'll answer other questions on my website listed in my profile
 
Mass is involved. Conceptually, greater mass means a tighter orbit. You can also check with dimensional analysis. The universal gravitational constant has units L3/M/T2. The right hand side of the equation in the original post, [itex]((4\pi^2)/(GM))r^3[/itex] has units T2, which agrees with the left-hand side.

To shade585:
You know you have to solve [itex]T^2 = ((4\pi^2)/(GM))r^3[/itex] for [itex]r[/itex]. How did you go about doing this? What did you use for the period of Mercury?
 
If I am understanding synchronous orbit correctly I am using the average rotational period of mercury which is 58.7 days.

(((T^2)(GM))/4pi^2))^1/3 = r I know i need to get unitsof meters in the end but can't figure out what to use for T in order to end up with just those units.
 
You have to use consistent units. For example, if you use 7.6159e-05 furlong3/fortnight2/stone as the value for G, you had better express the mass of Mercury in stones and its period in fortnights. The result will be in furlongs. Suppose instead you use the standard value for G, 6.673e-11 m3/s2/kg. What units do you think you need to use for the period in this case?
 
seconds?

The kg would be divided out. G(T) would leave m^3
 
Ok thanks for your help I figured it out.
 
You're welcome.
 

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