Question

2. Assume your mass is 102 kg and the acceleration due to gravity is 10m/s.
a. Use the formula for force F=ma to calculate your weight on Earth.
b. Make a prediction. What would happen to your weight if the Earth's
radius was reduced in half?
c. Let's mathematically prove whether or not you are correct. The Earth's
new radius is 3.185 x 106. Earth's mass is 5.98 x 1024kg. Your mass is
102 kg. What is the Gravitational force between you and this new
Earth (this is also your weight)? Where you correct?
Define Variables
Write equation and show all work.
Answer with
correct units.
m2=
Was your
prediction
correct? Y/N

Answer 1

a) Weight: 1020 N

b) New weight: 4080 N

c) Gravitational force: 4011 N

Explanation:

a)

In this problem, we calculate the weight of the person on Earth using the following equation:

$F=ma$

where:

m is the mass of the person

a is the acceleration due to gravity

In this problem, we have

m = 102 kg (mass)

$a=10 m/s^2$ (acceleration)

Substituting, we find the weight:

$F=(102)(10)=1020 N$

b)

The acceleration due to gravity is given by

$a=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

In this problem:

- The radius of the Earth is reduced in half: $R'=\frac{R}{2}$

- The mass of the Earth remains the same: $M'=M$

Therefore, the new value of the acceleration of gravity would be

$a'=\frac{GM'}{R'^2}=\frac{GM}{(R/2)^2}=4(\frac{GM}{R^2})=4a=4\cdot (10 m/s^2)=40 m/s^2$

And so, the new weight would be

$F'=ma'=(102)(40)=4080 N$

c)

The gravitational force between two objects is given by Newton's law of universal gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem:

$m_1 = 102 kg$ is the mass of the man

$m_2=5.98\cdot 10^{24} kg$ is the mass of the Earth

$r=3.185\cdot 10^6 m$ is the Earth's radius

Substituting, we find

$F=(6.67\cdot 10^{-11})\frac{(102)(5.98\cdot 10^{24})}{(3.185\cdot 10^6)^2}=4011 N$

Therefore, we see that this is approximately equal to the weight estimated in part 2.

Learn more about gravitational force:

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