Question

The formula for eccentricity, e, of an orbit is given below, where a is the length of the semi-major axis and b is the length of the major axis.
Solve the formula for eccentricity for the length of the semi-major axis.



The correct equation for the length of the semi-major axis is equation .

When the length of the major axis is 4 inches and the eccentricity of the orbit is , the length of the semi-major axis is inches (round to the nearest hundredth).

Reset
Submit

Answer 1

Answer:

Part 1) $a=\frac{b}{\sqrt{1-e^2}}$

Part 2) $a=4.62\ in$

Step-by-step explanation:

The correct question in the attached figure

Part 1)

we have the formula

$e=\sqrt{1-\frac{b^2}{a^2}}$

where

e the eccentricity

a is the length of the semi-major axis

b is the length of the major axis

solve for a

That means

Isolate the variable a

$e=\sqrt{1-\frac{b^2}{a^2}}$

squared both sides

$e^2=1-\frac{b^2}{a^2}$

subtract 1 both sides

$e^2-1=-\frac{b^2}{a^2}$

Multiply both sides by a^2

$(e^2-1)a^2=-b^2$

Divide both sides by (e^2-1)

$a^2=\frac{-b^2}{e^2-1}$

Rewrite

$a^2=\frac{-b^2}{-(1-e^2)}$

$a^2=\frac{b^2}{(1-e^2)}$

square root both sides

$a=\sqrt{\frac{b^2}{(1-e^2)}}$

simplify

$a=\frac{b}{\sqrt{1-e^2}}$

Part 2) we have

$b=4\ in\\e=1/2$

Solve for a

substitute in the formula of part 1)

$a=\frac{4}{\sqrt{1-(1/2)^2}}=4.62\ in$