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Kobotan [32]
3 years ago
15

Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

nal force on you?

Physics
2 answers:
Pachacha [2.7K]3 years ago
5 0
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
skelet666 [1.2K]3 years ago
3 0

The ratio of gravitational force of the sun to the earth's gravitational force is about 1 : 1580

\texttt{ }

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows :

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

radius of Earth = Re = 6.4 × 10⁶ m

mass of the Earth = Me = 6.0 × 10²⁴ kg

distance from Sun to Earth = Rs = 147 × 10⁹ m

mass of the Sun = Ms = 2.0 × 10³⁰ kg

<u>Asked:</u>

ratio of gravitational force of the sun and the earth = ?

<u>Solution:</u>

F_s : F_e = G \frac{ m M_s }{R_s^2} : G \frac{ m M_e }{R_e^2}

F_s : F_e = \frac{ M_s }{R_s^2} : \frac{ M_e }{R_e^2}

F_s : F_e = \frac{ 2 \times 10^{30} }{(147 \times 10^9)^2} : \frac{ 6.0 \times 10^{24} }{ (6.4 \times 10^6)^2}

F_s : F_e \approx 1 : 1580

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

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From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
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12(3.0)=m(6.0- \: 0)
This implies that

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You can also use the equation of linear motion,
v = u + at
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a = 2 {ms}^{ - 2}
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3 years ago
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soldi70 [24.7K]

Answer:

25N

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That should be right if im not dumb...

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At a distance r from a charge e on a particle of mass m the electric field value is  8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.

<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.

There are different ways where magnitude is used Magnitude of earthquake,  charge on an electron, force, displacement, Magnitude of gravitational force

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An ice skater accelerates backward for 5.0 second to final speed of 12m/s. If the acceleration backward was at a a rate of 1.5m/
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Given the following data;

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To find the initial velocity, we would use the first equation of motion.

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V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

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