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3 years ago

Find an exact expression for the orbital period T. Hint: Each planet feels two forces.

Mathematics

1 answer:

Otrada [13]3 years ago

6 0

Answer:

T^2 = (4π^2/GM)a^3

Step-by-step explanation:

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit” That’s Kepler’s third law. In other words, if you square the ‘year’ of each planet, and divide it by the cube of its distance to the Sun, you get the same number, for all planets.

(The other two are “the orbit of each planet is an ellipse with the Sun at a focus”, and “a line between a planet and the Sun sweeps out equal areas in equal times”.)

Copernicus, Kepler, and Newton dealt a one-two-three knockout blow to the idea – thousands of years old – that the Sun (and planets) moved around the Earth. Copernicus put the Sun at the center, Kepler modified Copernicus’ circular motions (and provided a simple, quantitative description of the actual motion), and Newton explained how it all worked (gravity).

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Write 120% as a fraction.

Olin [163]

120 percent=120/100=60/50=30/25=6/5, So it simplifies to 6/5.

5 0

3 years ago

Read 2 more answers

Subset of {a b c d e f} needs 64 subsets

spin [16.1K]

Answer:

Determine the power set of S, denoted as P:

Term Number Binary Term Subset

12 01100 {b,c}

13 01101 {b,c,e}

14 01110 {b,c,d}

15 01111 {b,c,d,e}

Step-by-step explanation:

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8 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Solve for b. (g+l)b=n

denis23 [38]

Answer:

b = ---------

(g+l)

Step-by-step explanation:

We need to isolate b. Starting from (g+l)b=n, we divide both sides by (g+l), obtaining

b = ---------

(g+l)

7 0

3 years ago

Represent the geometric series using the explicit formula. 3, −6, 12, −24, … f(n) = 3 ⋅ (2)(n−1) f(n) = 3 ⋅ (−2)(n−1) f(n) = f(n

mr Goodwill [35]

Answer:

$f(n)=(-2)(n-1)$

Step-by-step explanation:

Each term of this Geometric Series (3, −6, 12, −24, ...) can also be found through this explicit formula: $f(n)=(-2)(n-1)$ Because the $n^{th}$ term is found by the product of its common ratio "q", times its predecessor n-1. Where n, refers to the order of the term.

So let's test it, suppose we want to find the 4th term. We know the common ratio and the first term. Then we can write f(n) as:

$\\f(n)=3(-2)^{n-1}$

f(4)=f(3)*-2⇒-24=12*-2⇒-24=-24

The explicit formula is ok.

3 0

3 years ago