Respuesta :
b) The force with a distance of 150 km is 889 N
c) The force with a distance of 50 km is 8000 N
This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question.
Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as
F = (G M1 M2)/r^2
where
F = Force
G = gravitational constant
M1 = Mass 1
M2 = Mass 2
r = distance between center of masses for the two masses.
So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889.
Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits.
If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.
(b). The gravitational force between the objects when they are 150 km apart is [tex]\boxed{889\,{\text{N}}}[/tex].
(c). The gravitational force between the objects when they are 50 km apart is [tex]\boxed{8000\,{\text{N}}}[/tex] .
Further Explanation:
The gravitational force of attraction between the two bodies is given by the Newton’s Law of Gravitation. According to Newton’s law of Gravitation, the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The gravitational force is expressed mathematically as:
[tex]F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}[/tex]
Here, [tex]G[/tex] is the gravitational constant, [tex]{m_1}[/tex] is the mass of first body, [tex]{m_2}[/tex] is the mass of second body and [tex]r[/tex] is the distance between two bodies.
For two bodies kept at a distance of 100 Km , the gravitational force of attraction is 2000 N.
Substitute 200 N for [tex]F[/tex] and 100 Km for r in above equation.
[tex]\begin{aligned}2000=\frac{{G{m_1}{m_2}}}{{{{\left( {100\times {{10}^3}}\right)}^2}}}\hfill\\G{m_1}{m_2} = 2\times {10^{13}}\hfill\\\end{aligned}[/tex]
Part (b):
Now, the force experienced by the bodies when they are 150 Km apart is:
[tex]F = \dfrac{{G{m_1}{m_2}}}{{\left( {150 \times {{10}^3}}\right)}}[/tex]
Substitute [tex]2 \times{10^8}[/tex] for [tex]G{m_1}{m_2}[/tex] in above equation.
[tex]\begin{aligned}F&=\frac{{2 \times {{10}^{13}}}}{{{{\left( {150 \times {{10}^3}}\right)}^2}}}\\&= 888.9\,{\text{N}}\\&\approx {\text{889}}\,{\text{N}}\\\end{aligned}[/tex]
Thus, the gravitational force between the objects when they are 150 Km apart is [tex]\boxed{889\,{\text{N}}}[/tex].
Part (c):
Now, the force experienced by the bodies when they are 50 km apart is:
[tex]F =\Dfrac{{G{m_1}{m_2}}}{{\left( {50 \times {{10}^3}}\right)}}[/tex]
Substitute [tex]2 \times {10^8}[/tex] for [tex]G{m_1}{m_2}[/tex] in above equation.
[tex]\begin{aligned}F &=\frac{{2 \times {{10}^{13}}}}{{{{\left({50 \times {{10}^3}}\right)}^2}}}\\&= 8000\,{\text{N}}\\\end{aligned}[/tex]
Thus, the gravitational force between the objects when they are 50 Km apart is [tex]\boxed{8000\,{\text{N}}}[/tex].
Learn More:
1. Calculate the total force on the earth due to Venus, Jupiter, and Saturn https://brainly.com/question/2887352
2.Compare the surface area–to–volume ratios of Earth and Venus https://brainly.com/question/7227193
3.A rocket being thrust upward as the force of the fuel being burned https://brainly.com/question/11411375
Answer Details:
Grade: College
Subject: Physics
Chapter: Newton’s law of Gravitation
Keywords: Gravitation, newton’s law, force of attraction, 2000N, 100km, 150 km, 50 km, gravitational force, two spherical objects, 889 N, 8000 N.
