In celestial mechanics, the orbital speed of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ satisfies $v = \sqrt{\frac{G M}{r}}$, while the period of orbit is related by $T = \frac{2\pi r}{v}$. Eliminate the orbital speed $v$ by coupling these two literal relations and solve explicitly for the orbital radius $r$ in terms of the gravitational constant $G$, central mass $M$, and orbital period $T$.

In celestial mechanics, the orbital speed of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ satisfies $v = \sqrt{\frac{G M}{r}}$ , while the period of orbit is related by $T = \frac{2\pi r}{v}$ . Eliminate the orbital speed $v$ by coupling these two literal relations and solve explicitly for the orbital radius $r$ in terms of the gravitational constant $G$ , central mass $M$ , and orbital period $T$ .

Final answer

$r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$

Steps

Express the speed $v$ in terms of $r$ and $T$

To solve for the orbital radius $r$ in terms of $G$ , $M$ , and $T$ , we eliminate the intermediate variable $v$ using algebraic substitution and properties of exponents.

From the period formula: $T = \frac{2\pi r}{v}$

Multiply both sides by $v$ and divide by $T$ to isolate $v$ : $v = \frac{2\pi r}{T}$

Equate the two expressions for $v$

We are given the orbital speed formula: $v = \sqrt{\frac{G M}{r}}$

Substitute this into the expression from Step 1: $\sqrt{\frac{G M}{r}} = \frac{2\pi r}{T}$

Eliminate the radical by squaring both sides

Square both sides of the equation to clear the square root: $\left(\sqrt{\frac{G M}{r}}\right)^2 = \left(\frac{2\pi r}{T}\right)^2$

$\frac{G M}{r} = \frac{4\pi^2 r^2}{T^2}$

Common Pitfall: When squaring the fraction $\frac{2\pi r}{T}$ , be sure to square every factor in the numerator, including the coefficient: $(2\pi)^2 = 4\pi^2$ , not $2\pi^2$ .

Isolate $r$

Multiply both sides by $r$ to gather all powers of $r$ on one side: $G M = \frac{4\pi^2 r^3}{T^2}$

Multiply both sides by $T^2$ and divide by $4\pi^2$ : $r^3 = \frac{G M T^2}{4\pi^2}$

Take the cube root (raise both sides to the power of $\frac{1}{3}$ ) to solve explicitly for $r$ : $r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to square the factor of 2π when squaring both sides of the equation, leading to an incorrect coefficient of 2π² instead of 4π² in the denominator.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = $\frac{2\pi r}{\sqrt{\frac{GM}{r}}}$
Step 2
T^2 = $\frac{2\pi^2 r^2}{GM/r}$
Correct
T^2 = $\frac{4\pi^2 r^2}{GM/r}$
When squaring the numerator $2\pi r$ , the exponent must apply to every factor: $(2\pi r)^2 = 2^2 \cdot \pi^2 \cdot r^2 = 4\pi^2 r^2$ . You wrote $2\pi^2 r^2$ , omitting the square on the coefficient $2$ .
Step 3
T^2 = $\frac{2\pi^2 r^3}{GM}$ $\implies$ r^3 = $\frac{GM T^2}{2\pi^2}$
Correct
T^2 = $\frac{4\pi^2 r^3}{GM}$ $\implies$ r^3 = $\frac{GM T^2}{4\pi^2}$
You simplified the complex fraction and isolated $r^3$ correctly based on your previous line, but the result carried forward the incorrect coefficient $2\pi^2$ instead of $4\pi^2$ .
Step 4
r = $\sqrt[3]{\frac{GM T^2}{2\pi^2}}$
Correct
r = $\sqrt[3]{\frac{GM T^2}{4\pi^2}}$ = $\left(\frac{GM T^2}{4\pi^2}\right)$ ^{1/3}
You correctly took the cube root of both sides to isolate $r$ , but the expression still reflects the earlier coefficient mistake.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the expression for orbital velocity directly into the period equation.

You handled the division by a fraction in the denominator accurately to obtain $r^3$ .

You correctly solved for $r$ by taking the cube root of both sides in the final step.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Power of a Product Rule Misapplication

You squared the product $2\pi r$ as $2\pi^2 r^2$ rather than squaring each factor to get $4\pi^2 r^2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete distribution of the exponent over numerical and symbolic coefficients within a single product term $(ab)^n = a^n b^n$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Power of a Product Rule: For any real numbers $a, b$ and integer exponent $n$ , $(ab)^n = a^n b^n$ . Every constant and variable factor inside parentheses must be raised to the power.
Always write explicit parentheses around the entire group before squaring, expanding $(2 \cdot \pi \cdot r)^2 = 2^2 \cdot \pi^2 \cdot r^2 = 4\pi^2 r^2$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Square both sides of $T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}}$ accurately: $T^2 = \frac{(2\pi r)^2}{\frac{GM}{r}} = \frac{4\pi^2 r^2}{\frac{GM}{r}}$ .
Simplify the right-hand complex fraction: $T^2 = \frac{4\pi^2 r^3}{GM}$ .
Multiply both sides by $GM$ and divide by $4\pi^2$ : $r^3 = \frac{GM T^2}{4\pi^2}$ .
Take the cube root of both sides: $r = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}$ .

Final answer

$r = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}$

NEXT STEPS

Final Takeaways from Your Teacher

Write out grouped bases explicitly in parentheses like $(2\pi r)^2$ before squaring.
Verify each factor's power individually: check $2^2 = 4$ , $\pi^2$ , and $r^2$ .
Complete the three short practice exercises below to solidify literal exponent handling.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly simplifies the division by 1/r as multiplying by 1/r, reducing the power of r to r instead of r³.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = $\frac{2\pi r}{\sqrt{\frac{G M}{r}}}$
Step 2
T^2 = $\frac{4\pi^2 r^2}{\frac{G M}{r}}$
Step 3
T^2 = $\frac{4\pi^2 r}{G M}$
Correct
T^2 = $\frac{4\pi^2 r^3}{G M}$
When dividing by the fraction $\frac{G M}{r}$ , you should multiply by its reciprocal $\frac{r}{G M}$ . This gives $(4\pi^2 r^2) \cdot \frac{r}{G M} = \frac{4\pi^2 r^3}{G M}$ , rather than reducing the power of $r$ .
Step 4
r = $\frac{G M T^2}{4\pi^2}$
Correct
$r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$
Your algebraic isolation followed correctly from your previous line, but it inherited the missing cube power of $r$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the expression for $v$ into the period equation.

You squared every factor in the numerator and cleared the square root in the denominator properly.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Compound Fraction Division Error

You divided $r^2$ by $r$ instead of multiplying $r^2$ by the reciprocal of the denominator.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dividing a quantity by a fraction $\frac{A}{B}$ is equivalent to multiplying by its reciprocal $\frac{B}{A}$ . Here, dividing $r^2$ by $\frac{G M}{r}$ means multiplying by $\frac{r}{G M}$ , which increases the degree to $r^3$ instead of decreasing it to $r$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Reciprocal of a Denominator Fraction: Dividing by $\frac{a}{b}$ means multiplying by $\frac{b}{a}$ : $\frac{x}{\frac{a}{b}} = x \cdot \frac{b}{a}$ .
Always rewrite division by a fraction explicitly as multiplication by the reciprocal before combining terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite division by the denominator as multiplication by the reciprocal: $T^2 = 4\pi^2 r^2 \cdot \frac{r}{G M} = \frac{4\pi^2 r^3}{G M}$ .
Multiply both sides by $G M$ and divide by $4\pi^2$ : $r^3 = \frac{G M T^2}{4\pi^2}$ .
Take the cube root of both sides to isolate $r$ : $r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$ .

Final answer

$r = \left(\frac{G M T^2}{4\pi^2}\right)^{1/3}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice simplifying compound fractions by explicitly rewriting the division as multiplication by the reciprocal.
Check your final formulas using dimensional analysis or known physical laws (such as Kepler's Third Law, where $T^2 \propto r^3$ ).

OverviewACEpath Teacher's Diagnosis

03 The student takes a square root instead of a cube root when solving for r from the equation containing r³, leading to an incorrect root index.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
T = $\frac{2\pi r}{\sqrt{\frac{GM}{r}}}$
Step 2
T^2 = $\frac{4\pi^2 r^2}{\frac{GM}{r}}$ $\implies$ T^2 = $\frac{4\pi^2 r^3}{GM}$
Step 3
$r^3 = \frac{GMT^2}{4\pi^2}$
Step 4
$r = \sqrt{\frac{GMT^2}{4\pi^2}}$
Correct
r = $\sqrt[3]{\frac{GMT^2}{4\pi^2}}$ $\text{ or }$ r = $\left(\frac{GMT^2}{4\pi^2}\right)$ ^{1/3}
When solving $r^3 = A$ , the inverse operation of raising to the third power is taking the cube root (or raising to the power of $\frac{1}{3}$ ), not taking a square root.

WHAT YOU DID WELL

Compliments from Your Teacher

You substituted the velocity relation into the period equation correctly to eliminate $v$ .

You squared the entire numerator $(2\pi r)^2 = 4\pi^2 r^2$ without dropping the factor of $4$ .

You handled the division by a fraction correctly to combine the powers into $r^3$ .

You correctly isolated the variable term $r^3$ using inverse operations.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Root index mismatch when inverting powers

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You defaulted to writing a standard radical symbol $\sqrt{\dots}$ (index $2$ ) rather than specifying a cube root $\sqrt[3]{\dots}$ (or fractional exponent $\frac{1}{3}$ ) to invert $r^3$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Match Radical Index to Exponent: For any equation of the form $x^n = A$ , the inverse operation is $x = A^{1/n} = \sqrt[n]{A}$ .
Always check the exponent on the isolated variable before drawing a radical sign; if the power is $3$ , write an explicit index $3$ as $\sqrt[3]{\dots}$ or write the power as $(\dots)^{1/3}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Take the cube root of both sides (or raise both sides to the power of $\frac{1}{3}$ ):
$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} = \sqrt[3]{\frac{GMT^2}{4\pi^2}}$

Final answer

$r = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between inverting quadratic powers ( $x^2 \to \sqrt{x}$ ) versus cubic powers ( $x^3 \to \sqrt[3]{x}$ or $x^{1/3}$ ).
Practice using rational exponent notation $(\dots)^{1/3}$ when solving physics equations to prevent accidentally drawing a square root radical.

Lock it in

Question

What is the general expansion rule for $(c \cdot x)^n$ , where $c$ is a constant coefficient and $x$ is a variable?

Answer

$(c \cdot x)^n = c^n x^n$ . Every factor inside the product must be raised to the exponent $n$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to square the factor of 2π when squaring both sides of the equation, leading to an incorrect coefficient of 2π² instead of 4π² in the denominator.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student incorrectly simplifies the division by 1/r as multiplying by 1/r, reducing the power of r to r instead of r³.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student takes a square root instead of a cube root when solving for r from the equation containing r³, leading to an incorrect root index.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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