# The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is...

## Question:

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distance of {eq}1.86 \times 10^4 {/eq} km above the planet's surface, his weight changes to 24.31 N. What is the mass of planet X if Speff's mass is 75.0 kg?

## Gravitational Force:

The weight of an object at the surface of a planet is due to a force of attraction acting on the object toward the center of the planet that force is called the gravitational force. The gravitational force acting on a mass m at the surface of the planet of radius R and mass M can be mathematically expressed as follows:

{eq}F\ =\ G\cdot \dfrac{Mm}{R^2} {/eq}.

We are given:

• The weight of the spaceman Speff at the surface of the planet X is {eq}W_s\ =\ 389 \rm N{/eq}.
• The weight of the spaceman Speff at a height {eq}h\ =\ 1.86\times 10^4\ \rm km{/eq} from the surface of the planet X is {eq}W_h\ =\ 24.31 \rm N{/eq}.
• The mass of the spaceman Speff is {eq}m\ =\ 75.0\ \rm kg{/eq}.

We are asked to calculate the mass of Planet X. Therefore, we need to calculate the radius of the planet X to calculate the mass of the planet by using the weight of the spaceman as follows:

{eq}\begin{align} W_s\ &=\ mg_s\\[0.3 cm] g_s\ &=\ \dfrac{W_s}{m}\\[0.3 cm] &=\ \dfrac{389\ \rm N}{75.0\ \rm kg}\\[0.3 cm] &\approx \ 5.19\ \rm m/s^2\\[0.3 cm] \end{align} {/eq}

Now, we are also given:

{eq}\begin{align} W_h\ &=\ mg_h\\[0.3 cm] g_h\ &=\ \dfrac{W_h}{m}\\[0.3 cm] g_s\left (1-\dfrac{2h}{R}\right )\ &=\ \dfrac{24.31\ \rm N}{75.0\ \rm kg}\\[0.3 cm] \left (1-\dfrac{2h}{R}\right )\ &=\ \dfrac{0.324\ \rm m/s^2}{g_s}\\[0.3 cm] \left (1-\dfrac{2h}{R}\right )\ &=\ \dfrac{0.324\ \rm m/s^2}{5.19\ \rm m/s^2}\\[0.3 cm] \left (1-\dfrac{2h}{R}\right )\ &=\ 0.062\\[0.3 cm] \dfrac{2h}{3}\ &=\ 0.938\ \rm m\\[0.3 cm] R\ &=\ \dfrac{2h}{0.938\ \rm m}\\[0.3 cm] &=\ \dfrac{2\times 1.86\times 10^4\ \rm km}{0.938\ \rm m}\\[0.3 cm] &\approx \ 3.97\times 10^4\ \rm km\\[0.3 cm] \end{align} {/eq}

Now, the mass of the planet X can be calculated as follows:

{eq}\begin{align} g_s\ &=\ 5.19\ \rm m/s^2\\[0.3 cm] G\cdot \dfrac{M}{R^2}\ &=\ 5.19\ \rm m/s^2\\[0.3 cm] M\ &=\ (5.19\ \rm m/s^2)\times \dfrac{R^2}{G}\\[0.3 cm] &=\ (5.19\ \rm m/s^2)\times \dfrac{(3.97\times 10^7\ \rm m)^2}{6.67\times 10^{-11}\ \rm N\cdot m^2/kg^2}\\[0.3 cm] &\approx \ \boxed{\color{green}{1.23\times 10^{26}\ \rm kg}}\\[0.3 cm] \end{align} {/eq}

Gravitational Pull of the Earth: Definition & Overview

from

Chapter 15 / Lesson 17
31K

Earth's gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we'll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.