If a spacecraft has a high specific mechanical energy, what does this tell us about the size of the orbit? Why?
What is the specific mechanical energy, e, of an orbit with a semimajor axis of 42,160 km
Solution
Mechanical energy = kinetic energy + potential energy
ME = KE + PE
The SPECIFIC mechanical energy is just the ME divided by the spacecraft\'s mass:
SME = ME/m = (KE+PE)/m
(The reason they like to use SME instead of ME, is that it simplifies the equation a bit, explained later.)
Now let\'s work on the KE. Let\'s assume it\'s a circular orbit with radius R and speed v. You know that the centripetal force acting on an object in uniform circular motion is:
Fcentrip = mv²/R
AND, you know that the gravitational force acting on the spacecraft is:
Fgrav = GMm/R²
(where \"M\" = mass of earth, or whatever it\'s orbiting around)
Since Fgrav is the ONLY force acting on the craft, it IS the centripetal force:
Fgrav = Fcentrip
GMm/R² = mv²/R
Solve for v²:
v² = GM/R
So the craft\'s KE is:
KE = ½mv² = GMm/(2R)
Now let\'s work on the PE:
You learned in class that the PE of a spacecraft is, by convention, calculated relative to a \"zero-PE\" point which is defined to be infinitely far away (infinitely \"high\" from the planet\'s surface). That means, any real point is lower than zero, so the calculated PE is negative. You did the calculus to discover that the spacecraft\'s PE is in fact:
PE = ?GMm/R
So the mechanical energy is:
ME = KE+PE = GMm/(2R) ? GMm/R = ?GMm/(2R)
Now divide by \"m\" to Get the Specific Mechanical Energy:
SME = (?GMm/(2R)) / m = ?GM/(2R)
So you see that the SME is always negative; that means a \"high SME\" is less negative (closer to zero) than a \"low SME\". To relate that to the size of the orbit (\"R\"), ask yourself: To get a high SME (one that\'s less negative), should you increase or decrease \"R\"?
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If a spacecraft has a high specific mechanical energy- what does this.docx
1. If a spacecraft has a high specific mechanical energy, what does this tell us about the size of the
orbit? Why?
What is the specific mechanical energy, e, of an orbit with a semimajor axis of 42,160 km
Solution
Mechanical energy = kinetic energy + potential energy
ME = KE + PE
The SPECIFIC mechanical energy is just the ME divided by the spacecraft's mass:
SME = ME/m = (KE+PE)/m
(The reason they like to use SME instead of ME, is that it simplifies the equation a bit, explained
later.)
Now let's work on the KE. Let's assume it's a circular orbit with radius R and speed v. You
know that the centripetal force acting on an object in uniform circular motion is:
Fcentrip = mv²/R
AND, you know that the gravitational force acting on the spacecraft is:
Fgrav = GMm/R²
(where "M" = mass of earth, or whatever it's orbiting around)
Since Fgrav is the ONLY force acting on the craft, it IS the centripetal force:
Fgrav = Fcentrip
GMm/R² = mv²/R
Solve for v²:
2. v² = GM/R
So the craft's KE is:
KE = ½mv² = GMm/(2R)
Now let's work on the PE:
You learned in class that the PE of a spacecraft is, by convention, calculated relative to a "zero-
PE" point which is defined to be infinitely far away (infinitely "high" from the planet's
surface). That means, any real point is lower than zero, so the calculated PE is negative. You did
the calculus to discover that the spacecraft's PE is in fact:
PE = ?GMm/R
So the mechanical energy is:
ME = KE+PE = GMm/(2R) ? GMm/R = ?GMm/(2R)
Now divide by "m" to Get the Specific Mechanical Energy:
SME = (?GMm/(2R)) / m = ?GM/(2R)
So you see that the SME is always negative; that means a "high SME" is less negative (closer to
zero) than a "low SME". To relate that to the size of the orbit ("R"), ask yourself: To get a
high SME (one that's less negative), should you increase or decrease "R"?
