Calc 3 - Line integrals The mass M of the Sun is approximately 1.989 × 10° kg, and the mass m of the Earth is approximately5.972 x 104 kg. In appropriate rectangular coordinates with the Sun at the origin, the position functionp cos(t)r(1) = 1+e cos(t)'1+ € cos(t)p sin(t)0 The gravitational fieldGMG(x, y, z) =(x, y, z)(x² + y? + z?)3/2is the gravitational force per unit mass of a particle with mass M > 0 located at the origin whereG - 6.674 × 10¬1' is the universal gravitation constant.

Answered: The mass M of the Sun is approximately…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Calc 3 - Line integrals

The mass M of the Sun is approximately 1.989 × 10° kg, and the mass m of the Earth is approximately
5.972 x 104 kg. In appropriate rectangular coordinates with the Sun at the origin, the position function
p cos(t)
r(1) = 1+e cos(t)'1+ € cos(t)
p sin(t)
0 <t< r
describes the orbit C of the Earth from perihelion (the point closest to the sun) to aphelion (the point furthest from
the sun), where e z 0.167 and p = 1.496 × 10° km. Find the work done by the gravitational force of the Sun on
the Earth from perihelion to aphelion. Provide correct units for your answer. (Note that the gravitational force of the
Sun on the Earth is mG.)

The gravitational field
GM
G(x, y, z) =
(x, y, z)
(x² + y? + z?)3/2
is the gravitational force per unit mass of a particle with mass M > 0 located at the origin where
G - 6.674 × 10¬1' is the universal gravitation constant.

Expert Solution

Step 1

Let W be the work done by the gravitational force of the Sun on the earth from perihelion to aphelion, then

$W = \int F . d r$ ,

where r is the position function and F is the gravitational force of the Sun on the earth and this is equal to $m G$ , i.e. $F = m G$ , where

m is the mass of the Earth and $G = - \frac{G M}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}} <x, y, z>$ is the gravitational force per unit mass of a particle with mass M.

Hence, $F = m G = - \frac{G m M}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}} <x, y, z>$ .

Since r is function of t, and $r = <x, y, z>$ so $\frac{d r}{d t} = <\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}>$ .

And now

$F . \frac{d r}{d t} = - \frac{G m M}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}} <x, y, z> . <\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}> = - \frac{G m M}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}} (x \frac{d x}{d t} + y \frac{d y}{d t} + z \frac{d z}{d t}) . . . (1)$

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