Question

Why is the following situation impossible? A spacecraft is launched into a circular orbit around the Earth and circles the Earth once every hour.

Short answer

Spacecraft would need to be in orbit underground. That is not possible so the given condition is impossible.

Step-by-step solution

Given

$T=\text{Period = 1 hour}$

Concept

The escape velocity of the earth is the minimum required velocity for any object to be free from the gravitational field of the earth. It can be represented by the following formula:

$v_{escape\text{velocity}}=\sqrt{\frac{2MG}{R}}$

Where G is gravitational constant, M is the mass of the object, R is the distance from the centre of the mass.

Explain the reasoning

The orbital period is,

$$\begin{gathered} T^2=4{\pi }^2\frac{r^3}{GM}······\left(1\right) \\ \text{where},T=\text{Period = 1 hour = 3600}\text{s} \\ r=\text{orbital}\text{radius} \\ G=\text{Gravitational}\text{universal}\text{constant = 6.67}\times {\text{10}}^{-11}\text{N}·{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}} \\ M=\text{Mass}\text{of}\text{the}\text{plant = 5.97}\times {\text{10}}^{24}\text{kg} \end{gathered}$$

From Equation (1),

$$\begin{gathered} r=\sqrt[3]{\frac{{\left(3600\text{s}\right)}^2\left(6.67\times {10}^{-11}\text{N}·{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(5.97\times {10}^{24}\text{kg}\right)}{4{\pi }^2}} \\ =5.075\times {10}^6\text{m} \end{gathered}$$

Given that the radius of the earth is $6371\times {10}^6\text{m}$this would be impossible for a satellite to do with current technology.