Question

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The mass of Mars is 6.4multiply1023 kg and its radius is 3.4multiply106 m.)

(b) Calculate the magnitude of the gravitational force exerted by the human on Mars.

(c) For comparison, calculate the approximate magnitude of the gravitational force of this human on a similar human who is standing 4 meters away.

(d) What approximations or simplifying assumptions must you make in these calculations? (Note: Some of these choices are false because they are wrong physics!)

-Use the same gravitational constant in (a) and (b) despite its dependence on the size of the masses.

-Treat Mars as though it were spherically symmetric.

-Treat the humans as though they were points or uniform-density spheres.

-Ignore the effects of the Sun, which alters the gravitational force that one object exerts on another.

Answer 1

Answer:

a) $F=1.044\times 10^9\ N$

b)$F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

• mass of Mars, $M=6.4\times 10^{23}\ kg$

• radius of the Mars, $r=3.4\times 10^{6}\ m$

• mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

• Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

• Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

Answer 2

The gravitational forces required are given by Newton's law of

gravitation.

Correct responses:

(a) The magnitude of the gravitational force, F ≈ 1.005 × 10⁹ N

(b) 1.005 × 10⁹ N

(c) The gravitational force is approximately 2.669636 × 10⁻⁸ N

(d) Treat the humans as though they were points of uniform–density spheres

Methods used to arrive at the solutions

Mass of the human, m₁ = 80 kg

Mass of Mars, M₂ = 6.4 × 10²³ kg

Radius of Mars, r = 3.4 × 10⁶ m

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

(a) According to Newton's law of gravitation, we have;

$F = \mathbf{ \dfrac{G \cdot m_1 \cdot M_2}{r^2}}$

Therefore, we have;

$F = \dfrac{6.67408 \times 10^{-11} \, m^3 \cdot kg^{-1} \cdot s^{-2} \times 80 \, kg \times 6.4 \times 10^{23} \, kg}{3.4 \times 10^6 \, m} \approx \mathbf{ 1.005 \times 10^9 \, N}$

• The magnitude of the gravitational force exerted by Mars on the human, F ≈ 1.005 × 10⁹ N

(b) According to Newton's third law of motion, action and reaction are

equal and opposite, therefore;

• The force exerted by Mars on the human = The force exerted by the human on Mars ≈ 1.005 × 10⁹ N

(c) The gravitational force between the two humans is therefore;

$F = \dfrac{6.67408 \times 10^{-11} \times 80 \times 80}{4^2} \approx \mathbf{ 2.669632 \times 10^{-8}}$

The gravitational force between the two humans is therefore;

• Force between two humans, F ≈ 2.669632 × 10⁻⁸ N

(d) The distance between the two bodies in the law of gravitation, is  the distance between their center of mass

Given that the center of mass of the humans vary, we have;

• The simplifying assumption is to treat the humans as though they were points or uniform-density spheres

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