Mathematics, 22.06.2019 13:20 Annabeans1105

The equation t^2=a^3 shows the relationship between a planet’s orbital period, t, and the planet’s mean distance from the sun, a, in astronomical units, au. if the orbital period of planet y is twice the orbital period of planet x, by what factor is the mean distance increased?

Answers: 2

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Answers

Answer from: madisyn13

2^3/2 or option D

Step-by-step explanation:

I just took a test with this question on it and got perfect marks so I know it's right.

Answer from: smcculleymcculley

2^3/2

Step-by-step explanation:

The question is on formulae variation

Given T²=A³the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

Making T subject of the formulae

T²=A³square root both sides

T= √A³ OR (A³)^1/2

if the orbital period of planet Y is twice the orbital period of planet X then,

Y=2T

Y=2× √A³

Y=2×(A³)^1/2

Applying the laws of indices

Y=2×(A)^(3×1/2)

Y=2×(A)^3/2

Compare

A^3/2 and 2A^3/2

The mean distance increased by 2^3/2

Answer from: jazmineweb234

$2\sqrt2$ or 2.828

Step-by-step explanation:

The given equation is

T^2=A^3

Let the mean distance from the sun of planet X is A then

mean distance from the sun of planet Y is 2A.

Therefore, we have

For planet X-

T_x^2=A^3....(i)

For planet Y-

$T_y^2=(2A)^3\\\\T_y^2=8A^3....(ii)$

Divide equation (i) and (ii)

$\frac{T_x^2}{T_y^2}=\frac{A^3}{8A^3}\\\\\frac{T_x^2}{T_y^2}=\frac{1}{8}\\\\\frac{T_x}{T_y}=\frac{1}{2\sqrt2}\\\\T_y=2\sqrt2T_x$

Therefore, we can conclude that orbital period is increase by a factor of $2\sqrt2$ or 2.828

Answer from: levicorey846

Answer with explanation:

The Equation which is relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU.

→ T^2=A^3

For Planet , X

$A_{1}=x AU\\\\(T_{1})^2=x^3$

--------------------------------------(1)

And, For Planet ,Y

$A_{2}=2 x AU\\\\(T_{2})^2=(2 x)^3$

-----------------------------------------(2)

Dividing equation (1) by (2)

$\frac{(T_{1})^2}{(T_{2})^2}=\frac{x^3}{(2x)^3}$

$[\frac{(T_{1})}{(T_{2})}]^2=\frac{x^3}{8 x^3}\\\\\frac{(T_{1})}{(T_{2})}=\sqrt{\frac{1}{8}}\\\\ T_{2}={2\sqrt{2}}{T_{1}}$

→Orbital Period of Planet Y increases by, 2√2 times =2 ×1.414=2.828 times than planet X.

Answer from: destinyaus14

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

Answer from: ike9264

Given the equation:

T^2 =A^3

shows the relationship between a planet's orbital period T and the planet's mean distance from the sun, A in Astronomical units

then:

For planet X:

Orbital period is:

$T_{X} = (A)^{\frac{3}{2}}$ .....[1]

As per the statement:

If planet Y is k times the mean distance from the sun as planet X.

⇒ A planet Y= kA mean distance from the sun as planet X.

then orbital period of Planet Y is:

$T_{Y} = (kA)^{\frac{3}{2}}=k^{\frac{3}{2}}\cdot (A)^{\frac{3}{2}}$ ....[2]

Divide equation [2] by [1] we have;

$\frac{T_{Y}}{T_{X}} = \frac{k^{\frac{3}{2}}\cdot (A)^{\frac{3}{2}}}{A^{\frac{3}{2}}}$

Simplify:

$\frac{T_{Y}}{T_{X}} =k^{\frac{3}{2}}$

$T_{Y} =k^\frac{3}{2} T_X$

Therefore, the orbital period is increased by factor $k^\frac{3}{2}$

Answer from: josetheghost6313

Hello, Your answer is below.

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3. Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then A(Y) = 2A(X) [T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3 But [T(X)]^2 = [A(X)]^3 Thus [T(Y)]^2 = 2^3[T(X)]^2 [T(Y)]^2 / [T(X)]^2 = 2^3 T(Y) / T(X) = 2^3/2 Therefore, the orbital period increased by a factor of 2^3/2.

Thank You.

Your friend Papaguy.

Answer from: ylimekasos311

Hi Nh803158,

Your Question:

The equation T^2 = A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If the orbital period of planet Y is twice the orbital period of planet X, by what factor is the mean distance increased?

D 2^3/2

Answer from: naomicervero

So, let planet X's orbital period be T and its mean distance from the sun be A. Also let planet Y's orbital period be T_1, so that means if planet Y's mean distance from the sun were double that of planet X:

$Z^2=(2A)^3\\ Z^2=8A^3\implies\\ Z^2=8T^2 \implies\\ Z=2\sqrt{2}T$

Which means that the orbital period in planet Y is increased by a factor of $2\sqrt{2}$

Answer from: Sariyahhall1

The orbital increased by factor $2^{\frac{3}{2}}$ .Hence, correctander is option D.

Step-by-step explanation:

Given; T^2=A^3

Distance of the planet X from the sun = d

Time-period of the planet X:

T^2=d^3 ..(1)

Distance of the planet Y from the sun = d'= 2d

T'^2=d'^2=(2d)^3=8d^3 ..(2)

On dividing (1)and (2).

$\frac{T'}{T}=\frac{8d^3}{d^3}=8=2\sqrt{2}=2^1\times 2^{\frac{1}{2}}=2^{\frac{3}{2}}$

The orbital increased by factor $2^{\frac{3}{2}}$ .Hence, correctander is option D.

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