Question

At the surface of Jupiter’s moon Io, the acceleration due to gravity is $g=1.81{\text{\hspace{0.17em}m/s}}^{\text{2}}$. A watermelon weighs 44.0 N at the surface of the earth.

(a) What is the watermelon’s mass on the earth’s surface?

(b) What would be its mass and weight on the surface of Io?

Short answer

(a) The watermelon weighing 44 N on Earth has mass 4.49 kg on the Earth's surface.

(b) The mass and weight of the watermelon on the surface of Io are 4.49 kg and 8.13 Nrespectively.

Step-by-step solution

Given data

The acceleration due to gravity on Io is

$g_I=1.81{\text{\hspace{0.17em}m/s}}^{\text{2}}$

The weight of the watermelon is

$W=44\text{\hspace{0.17em}N}$

The acceleration due to gravity on Earth is

$g=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$

Gravitational force

The downward gravitational force on a body of mass $m$ is

$F_g=mg$ …(1)

Weight and mass of watermelon at different locations

From equation (1), the mass of the watermelon on Earth is

$\begin{array}{c}m=\frac{W}{g}\\ =\frac{44\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}\\ =4.49\times \left(\frac{1}{{\text{m/s}}^{\text{2}}}\right)\times \left(1\text{\hspace{0.17em}N}\times \frac{1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}}{1\text{\hspace{0.17em}N}}\right)\\ =4.49\text{\hspace{0.17em}kg}\end{array}$

The mass if $4.49\text{\hspace{0.17em}kg}$.

The mass of an object is same everywhere. Thus, the mass of the watermelon on Io is also 4.49 kg.

From equation (1), the weight of the watermelon on Io is

localid="1662114432519" $\begin{array}{c}{W}_I=m{g}_I\\ =4.49\text{\hspace{0.17em}kg}\times 1.81{\text{\hspace{0.17em}m/s}}^{\text{2}}\\ =8.13\times \left(1{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\times \frac{1\text{\hspace{0.17em}N}}{1{\text{\hspace{0.17em}kgm/s}}^{\text{2}}}\right)\\ =8.13\text{\hspace{0.17em}N}\end{array}$

Thus, the required weight is 8.13 N.