Question

Suppose a particular simple pendulum of length l_0 has a period of T_0 for small oscillations here on Earth.

(a) Suppose the original pendulum is set up with the same length l_0 on the surface of a new planet, Uzaz, and it oscillates with a period 5/4 as long as the period of Earth. Given the gravitational acceleration here on Earth is 9.8 m/s^2, what is the gravitational acceleration (g)(v) at the surface on Uzaz?

(b) If the radius of Uzaz is known to be 3.5x10^6 m, what must be the mass of Uzaz?

(c) Suppose Uzaz rotates once every 18.5 hours. At what altitude above the surface of Uzaz must a satellite orbit in order to maintain synchronicity with a point on the surface of Uzaz?

(d) Suppose your spaceship (8x10^5 N) enters into elliptical orbit around Uzaz. At perihelion, your ship is (9x10^6 m) from Uzaz, and aphelion it is (1.6x10^7 m) from Uzaz.

Find the following for part (d):

i. the semimajor axis of your ship's orbit.

ii. the eccentricity of your ship's orbit.

iii. the speed of your ship at perihelion and aphelion.

iv. the period of your ship's orbit.

Answer 1

Given: Length of the simple pendulum on Earth=

Time period of the simple pendulum on Earth =

a) Now, the general expression for the time period of a Simple Pendulum is given as:   

So, Time period of the simple pendulum on Earth,

   ...............(1) since gravitational acceleration on Earth is 9.8 m/s^2

Time period of the simple pendulum on Uzaz is

....................(2) Since, time period on Uzaz is 5/4 times the period of Earth, is the gravitational acceleration on Uzaz, remains same on Uzaz

Dividing (2) by (1) gives,

Squaring both sides,

So, gravitational acceleration of Uzaz is 6.272 m/ s^2

b) The formula for gravitational acceleration for any planet is given by:-

where G=Gravitational constant=6.673 x 10^-11 Nm²/kg²

M= mass of the planet in Kg, r = radius of the planet.

substituting the values, = 6.272 m/s^2, r = 3.5 x 10^6 m, G = 6.673 x 10^-11 Nm²/kg²

Kg

c)