Respuesta :
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 kg
Distance between centers of sun and earth = 149.6×10⁹ 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
Ratio = 6.04
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 kg
Distance between centers of sun and earth = 149.6×10⁹ 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
Ratio = 6.04
The ratio of gravitational force of the sun to the earth's gravitational force is about 1 : 1580
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Further explanation
Newton's gravitational law states that the force of attraction between two objects can be formulated as follows :
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
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Given:
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
Asked:
ratio of gravitational force of the sun and the earth = ?
Solution:
[tex]F_s : F_e = G \frac{ m M_s }{R_s^2} : G \frac{ m M_e }{R_e^2}[/tex]
[tex]F_s : F_e = \frac{ M_s }{R_s^2} : \frac{ M_e }{R_e^2}[/tex]
[tex]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}[/tex]
[tex]F_s : F_e \approx 1 : 1580[/tex]
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
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Answer details
Grade: High School
Subject: Physics
Chapter: Gravitational Fields
