A satellite moves in a circular orbit of radius 2R around a spherical planet of radius R a… — JAMB UTME

A satellite moves in a circular orbit of radius $2R$ around a spherical planet of radius $R$ and mass $M$ . It is subsequently shifted to a larger circular orbit of radius $8R$ . What is the ratio of its initial orbital speed to its new orbital speed?

A
$4 : 1$
$2 : 1$ Answer
C
$1 : 2$
D
$1 : 4$

Answer

The ratio of the initial orbital speed to the new orbital speed is

2 : 1

Orbital velocity varies inversely with the square root of orbital radius (

v \propto \frac{1}{\sqrt{r}}

). Moving from radius

2R

8R

decreases the speed by a factor of

\sqrt{8/2} = 2

. Therefore, the ratio of initial speed to new speed is

2 : 1

Step-by-Step Solution

Write the formula for orbital velocity

v = \sqrt{\frac{GM}{r}}

, where

G

is the gravitational constant,

M

is the mass of the planet, and

r

is the orbital radius.

Orbital speed is determined by equating gravitational force to centripetal force.

Set up expressions for initial and final orbital speeds

Initial speed

v_1 = \sqrt{\frac{GM}{2R}}

and final speed

v_2 = \sqrt{\frac{GM}{8R}}

Substitute the given orbital radii

r_1 = 2R

and

r_2 = 8R

into the formula.

Calculate the ratio

v_1 / v_2

\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{GM}{8R}}} = \sqrt{\frac{8R}{2R}} = \sqrt{4} = 2

Simplifying the square root fraction gives the exact ratio.

Key Concept

Orbital Velocity and Inverse Square Law Relations

Estimated Time:1m 15s

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