Question

A planet is in circular orbit around the Sun. Its distance from the Sun is four times the average distance of Earth from the Sun. The period of this planet, in Earth years, is:

Answer 1

Answer: B

Explanation: 8 Q1:

A planet is in circular orbit around the Sun. Its distance from the Sun is four times the

average distance of Earth from the Sun. The period of this planet, in Earth years, is :

A) 4

B) 8

C) 16

D) 32

E) 2

Ans:

B

Q2:

A satellite orbits a planet of unknown mass in a circle of radius 2.0x107

Ans:

F = mv2

r

⇒

1

2 rf = 1

2 mv2 = K

m. The

magnitude of the gravitational force on the satellite form the planet is F = 76.0 N.

What is the kinetic energy of the satellite in this orbit?

K = 1

2

rF = 1

2 (2 × 107)(76) = 7.6 × 108 J

Q3:

A 22.4 kg satellite has a circular orbit with a period of 3.81 h and a radius of

8.0 x 106

Ans:

T2 = �

4π2

GM� r3

(3.81 × 3600)2 = � 4π2

6.67 × 10−11 × M� (8 × 106)3

M = 1.6 × 1024 kg

Answer 2

Answer:

42 m/a

Explanation:

It’s just that I can’t explain