A satellite revolves around a planet in a circular orbit of radius 1.0 \times 10^4 \text{ … — JAMB UTME

A satellite revolves around a planet in a circular orbit of radius $1.0 \times 10^4 \text{ km}$ with an orbital period of $12 \text{ hours}$ . Calculate the orbital period, in hours, of a second satellite orbiting the same planet in a circular path of radius $4.0 \times 10^4 \text{ km}$ .

Answer: 96 hours

Answer

96 hours

According to Kepler's Third Law (

T^2 \propto r^3

), the orbital period

T

scales with radius

r

T \propto r^{3/2}

. Increasing the orbital radius by a factor of 4 increases the period by a factor of

4^{3/2} = 8

. Multiplying the original period of 12 hours by 8 yields 96 hours.

Step-by-Step Solution

Set up Kepler's Third Law equation relating orbital period and orbital radius.

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

Kepler's Third Law states that the square of the orbital period of a body in circular orbit is directly proportional to the cube of the radius of its orbit.

Substitute the given orbital radii and evaluate the scaling factor.

\frac{r_2}{r_1} = \frac{4.0 \times 10^4 \text{ km}}{1.0 \times 10^4 \text{ km}} = 4

Simplifying the ratio of the two orbital radii gives a factor of 4 increase in radius.

Calculate the period multiplier by taking the ratio to the power of 3/2.

4^{3/2} = (\sqrt{4})^3 = 2^3 = 8

Taking

T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2}

shows the period scales by a factor of 8.

Multiply the initial orbital period by the scaling factor to find the final answer.

T_2 = 12 \text{ hours} \times 8 = 96 \text{ hours}

Multiplying the baseline period of 12 hours by 8 yields the new orbital period.

Key Concept

Kepler's Third Law of Planetary Motion

Estimated Time:1m 30s

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