American artificial planet Pioneer IV

[roger]

hi

Im stuck on how to do this question ...can anyone give me any hints :

The American artificial planet Pioneer IV reached its closest approach to the sun 91.7 million mi on march 17 1959.
At aphelion it will be 106.1 million mi from the sun.

What is:
a.) the speed at which it passed perihelion

b.) its sidereal period

thanks


Roger

 

[futb0l]

Use the conservation of mechanical energy equation, remember that capital M is equal to the mass of the sun.

<br /> E_{total} = \frac{1}{2}mv_p^2 - \frac{mMG}{r} = -\frac{mMG}{2a}<br />

a is equal to the semi major axis and 2a is equal to the distance from perihilion + the distance from the aphelion.

The period can be calculated using the following equation:

<br /> T^2 = \frac{4\pi^2a^3}{MG}<br />

;)

 

[wais]

, I can definitely give you some hints on how to approach this question. First, let's define some terms:

- Perihelion: This is the point in an object's orbit where it is closest to the sun.
- Aphelion: This is the point in an object's orbit where it is farthest from the sun.
- Sidereal period: This is the time it takes for an object to make one full orbit around the sun, relative to the stars.

Now, let's use some formulas to help us solve for the missing information:

a.) To find the speed at which Pioneer IV passed perihelion, we can use the formula for orbital velocity:

V = √(GM(2/r - 1/a))

- V represents the velocity
- G is the gravitational constant (6.674 x 10^-11 m^3/kg/s^2)
- M is the mass of the sun (1.989 x 10^30 kg)
- r is the distance from the sun at perihelion (91.7 million mi, which converts to 1.4722 x 10^11 m)
- a is the semi-major axis (half of the longest diameter of the elliptical orbit, which is equal to the average distance from the sun) (98.9 million mi, which converts to 1.5908 x 10^11 m)

Plugging in these values, we get:

V = √((6.674 x 10^-11)(1.989 x 10^30)(2/(1.4722 x 10^11) - 1/(1.5908 x 10^11)))

V = √(2.625 x 10^20 - 1.257 x 10^20)

V = √(1.368 x 10^20) = 1.170 x 10^10 m/s

So, the speed at which Pioneer IV passed perihelion was approximately 11.7 billion meters per second.

b.) To find the sidereal period, we can use the formula:

T = 2π√(a^3/GM)

- T represents the period
- a is the semi-major axis (1.5908 x 10^11 m)
- G is the gravitational constant (6.674 x 10^-11 m^3/kg/s^2)
- M is