Question

1.Zero, a hypothetical planet, has a mass of 4.9 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

2.In deep space, sphere A of mass 100 kg is located at the origin of an x axis and sphere B of mass 43 kg is located on the axis at x = 1.7 m. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as B is released? (b) What is the kinetic energy of B when it has moved 1.0 m toward A?

3.A satellite is placed in a circular orbit about Earth with a radius equal to 24% the radius of the Moon's orbit. What is its period of revolution in lunar months? (A lunar month is the period of revolution of the Moon.)

4.A 19.0 kg satellite has a circular orbit with a period of 3.70 h and a radius of 6.50 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 2.20 m/s2, what is the radius of the planet?

5.A satellite of mass 281 kg is in a circular orbit of radius 6.79 x106 m around a planet. If the period is 8090 s, what is the mechanical energy of the satellite?

6.A planet requires 320 (Earth) days to complete its circular orbit around its sun, which has a mass of 5.0 x 1030 kg. What are the planet's (a) orbital radius and (b) orbital speed?

Answer 1

1. Given,

The mass of zero, M = 4.9 x 10²³ kg

The radius of zero, R = 3.3 x 10⁶ m

The mass of a probe, m = 10 kg

(a)

Given, r = 4 x 10⁶ m

By using conservation of energy

Gravitational PE at surface + KE at surface = Gravitational PE + KE   at the given point

- GMm/R + KE = - GMm/r + KE

- (6.67 x 10^-11) ( 4.9 x 10²³ ) (10) / (3.3 x 10⁶ ) + 5 x 10⁷= - (6.67 x 10^-11) (4.9 x 10²³ ) (10) / (4 x 10⁶ ) + KE

K.E = (1 / (3.3 x 10⁶ )) - (1 / (4 x 10⁶ )) +  5 x 10⁷

   = 5 x 10⁷ J

(b)

By using the conservation of energy

Gravitational PE at surface + KE at surface = Gravitational PE at the given point

- GMm/R + KE = - GMm/r + KE

- (6.67 x 10^-11) (4.9 x 10²³ ) (10)/(3.3 x 10⁶ ) + KE = - (6.67 x 10^-11) (4.9 x 10²³ ) (10)/(8 x 10⁶ )

K.E = 7.01 x 10⁷ J