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zepelin
12 days 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:
kicyunya [1K]12 days ago
5 0
To find the gravitational force exerted on you by the Earth, we start with your weight estimated at around 60 kg.

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

Thus, 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 calculate the gravitational force due to the Sun by substituting,
m₁ = 1.989 × 10³⁰ kg and the distance between the centers of the Sun and Earth as 149.6 × 10⁹ m.
This gives us,
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

The ratio is then calculated as 0.356 N / 589.18 N, yielding Ratio = 6.04.
inna [987]12 days ago
3 0

The gravitational force ratio between the sun and the earth stands at approximately 1: 1580

\texttt{ }

Further explanation

According to Newton's law of gravitation, the force attracting two bodies can be represented mathematically as follows:

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

F = Gravitational Force ( Newtons )

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

m = Mass of the Object ( kg )

R = Separation Distance ( m )

Let’s solve this now!

\texttt{ }

Given:

Earth's radius = Re = 6.4 × 10⁶ m

Earth's mass = Me = 6.0 × 10²⁴ kg

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

Sun's mass = Ms = 2.0 × 10³⁰ kg

Requested:

what is the ratio of the gravitational forces of the sun to the earth?

Solution:

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{ }

Learn more

  • Consequences of Gravity:
  • Gravity's Effect on Objects:
  • Gravity Acceleration:

\texttt{ }

Answer details

Level: High School

Field: Physics

Chapter: Gravitational Fields

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A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
kicyunya [1025]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

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3 0
8 days ago
The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un
serg [1198]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation    x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

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6 0
1 day ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Softa [913]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Reservoir pressure = 10 atm

T_1 = Reservoir temperature = 300 K

P_2 = Exit pressure = 1 atm

T_2 = Exit temperature

R_s = Specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

Assuming isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

Flow density at exit measures 2.2721 kg/m³

4 0
11 days ago
The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top
Keith_Richards [1034]
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6 days ago
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serg [1198]

Response:

45cm

Clarification:

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Using the mirror formula to derive the object distance;

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Where f denotes the focal length, u indicates the object distance, and v represents the image distance.

Given f = 30cm, and v = 2u (The formed image is double the size of the pencil)

Plugging these values into the formula to solve for u yields;

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By cross-multiplying, we obtain;

2u = 90

Dividing both sides by 2;

2u/2 = 90/2

u = 45cm

The object's distance from the mirror measures 45cm

3 0
17 days ago
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