As mentioned on page \(13,\) the MCO was lost because of a mix-up in the units used to calculate the force needed to correct its trajectory. Ground-based computers generated the force correction file. On September \(29,1999,\) it was discovered that the forces reported by the ground-based computer for use in MCO navigation software were low by a factor of \(4.45 .\) The erroneous trajectory brought the MCO \(56 \mathrm{km}\) above the surface of Mars; the correct trajectory would have brought the MCO approximately \(250 \mathrm{km}\) above the surface. At \(250 \mathrm{km},\) the MCO would have successfully entered the desired elliptic orbit. The data contained in the force correction file were delivered in lb-sec instead of the required SI units of newton-sec for the MCO navigation software. The newton is the SI unit of force and is described in Appendix B. The British Engineering (gravitational) system uses a pound (lb) as a unit of force and \(\mathrm{ft} / \mathrm{s}^{2}\) as a unit of acceleration. In turn, the pound is defined as the pull of Earth on a unit of mass at a location where the acceleration due to gravity is \(32.174 \mathrm{ft} / \mathrm{s}^{2} .\) The unit of mass in this case is the slug, which is \(14.59 \mathrm{kg}\). Thus, BE unit of force \(=1\) pound \(=(\text { slug })\left(\mathrm{ft} / \mathrm{s}^{2}\right)\) Use this information to confirm that BE unit of force \(=4.45 \times\) SI unit of force 1 pound \(=4.45\) newton

Step by step solution

01

Identify the quantities given and the units that need conversion

The given problem mentions that 1 pound in the BE system equates to a slug multiplied by \( \frac{ft}{s^2} \), where a slug is \( 14.59 kg \). It also states that at a location where the acceleration due to gravity is \( 32.174 \frac{ft}{s^2} \), the Earth would pull a unit mass (slug) with a force of 1 pound. The aim is to confirm whether 1 pound equals to 4.45 newtons.

02

Write down the definition of a Newton (SI unit of force)

A newton is defined as the amount of force required to accelerate a one-kilogram mass by one meter per second squared. This can be written as: \(1 N = 1 \frac{kg \cdot m}{s^2}\).

03

Convert a pound (BE unit of force) to SI units

This step involves multiple subsets of conversion: Convert the slug to kilogram: \(1 slug = 14.59 kg\). Convert feet to meter: \(1 ft = 0.3048 m\). Substitute these conversion factors and plug in the given value of acceleration due to gravity into the equation for 1 pound (which equals to slug multiplied by \(\frac{ft}{s^2}\)): \(1 pound = 14.59 kg \cdot 32.174 (0.3048 \frac{m}{s^2}) = 4.45 \frac{kg \cdot m}{s^2}\). This is essentially 4.45 Newtons.

04

Confirm the conversion

As calculated in the previous step, a pound (BE unit of force) is equivalent to 4.45 Newtons (SI unit of force). Hence the conversion is confirmed.

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