For a planet orbiting the sun, r p is the distance from the sun to the perihelion and r a is the distance from the sun to the aphelion. What is a planet’s orbital eccentricity if r p is equal to 0.45r a?
A. 1.45
B. 0.55
C. 0.38
D. 2.64

Where is the sun located on Earth’s elliptical orbit?
A. at the center of the two foci
B. at a focus
C. at the aphelium
D. at the perihelium

A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to an orbital radius of 4r?
A. 1/4v
B. 4v
C. 1/2v
D. 2v

1) 0.38 (C)

2) At a focus (B)

3) (1/2)v (C)

Explanation:

1) Eccentricity, e = (Ra - Rp)/(Ra +Rp)

Let r a = Ra, rp = Rp

Rp = 0.45Ra

e = (Ra - 0.45Ra)/(Ra + 0.45Ra)

e= 0.55Ra/1.45Ra

e = 55/145

e= 0.38

2) The sun is located at a focus on Earth’s elliptical orbit:

The different location in the Earth’s orbit of the Sun is determined by the perihelion and aphelion.

The perihelion is the point on earth's orbit closest to the sun while aphelion is the point on earth's orbit that is far away from the sun.

The Sun tends a little bit towards one side at a point known as the focus of an ellipse.

This out of alignment causes the planet to move either closer to or further away from the Sun every orbit.

3) initial velocity = v

radius = r

v = √(GM/r)

v = (√GM/√r)

From the expression above, Velocity is directly proportional to the inverse of square root of radius

v = 1/√r

If radius become 4r, the new velocity (V):

V = 1/√4r

V = 1/(2√r) = 1/2(1/√r)

Recall, v = 1/√r

V = (1/2)(v)

V = v/2

Hence velocity decreases as r increases.

New velocity = v/2

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.