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MatroZZZ [7]
2 years ago
8

Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth’s orbit remained the same as it i

s now, then compared with the gravitational force between Earth and the sun, the gravitational force between Earth and the star would be—
half as much
the same
two times as much
four times as much
Physics
1 answer:
Vaselesa [24]2 years ago
7 0

Answer:

Two times as much

Explanation:

The equation for gravitational force is: Fg = GMm/r^2 with G being the universal gravitational constant.

So to make things easier we'll set r equal to 1 since it's a constant as well as G.

Then we're left with Fg=Mm with M being the mass of the sun and m being the mass of the earth.

So if m is constant and supposedly equals 1 then Fg=M so Fg is proportional to M therefore if M doubles then Fg doubles.

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Hello!

We can use the kinematic equation:
a = \frac{v_f - v_i}{t}

a = acceleration (m/s²)

vf = final velocity (45 m/s)
vi = initial velocity (25 m/s)

t = time (5 sec)

Plug in the givens:
a = \frac{45-25}{5} = \frac{20}{5} = \boxed{4 m/s^2}

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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
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Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

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