Ask question

Ask question

All categories

3 years ago

12

There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl

anet and the moon is 4.28 × 105 km, the orbital period of the moon is 28.9 days, and the radius of the moon is 1700 km. What is the angular momentum of the moon about the planet? Answer in units of kg m2 /s.

2 answers:

kaheart [24]3 years ago

7 0

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28 × 10⁵km

Orbital period T = 28.9days

T = 28.9 × 24 × 60 × 60

= 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

$L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s$

lorasvet [3.4K]3 years ago

6 0

Answer:

2.4094 × $10^{14}$ kg $m^{2}$ /sec

Explanation:

Mass of moon = 9.58 × $10^{22}$ kg

Radius of moon = 1700 km

Time period of moon = 28.9 days =2,496,960 sec

Angular momentum;

L=mvr --- (1)

where v= velocity

Velocity of moon =2 π r / T

Put in above;

==> L = m (2 π r) r / T

==> L = 2 π m $r^{2}$ / T

Putting data values;

==> L = 2×3.14×9.58 × $10^{22}$ × 2890000000 ÷ 2,496,960

==> L = 2.4094× $10^{22}$ × $10^{-8}$

==> L = 2.4094 × $10^{14}$ kg $m^{2}$ /sec

You might be interested in

Which of the following is NOT a characteristic of an inner planet?. . A.. rocky. . B.. solid surface. . C.. near the sun. . D..

sveticcg [70]

Being made mostly of gas is NOT a characteristic of an inner planet. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

3 0

3 years ago

Read 2 more answers

During which time does Jamal have the greatest momentum?

Yuliya22 [10]

Answer:

Momentum Packet Answer KEY - Science Online

YOU WILL SEE THE PDF

4 0

3 years ago

What is the biggest barrier to the use of renewable energy in the United States?

storchak [24]

The biggest barrier to the use of renewable energy in the United States is the Citizen opposition to negative environmental impact.
So, the answer is B.

7 0

3 years ago

A roller coaster is going through a loop with a circular radius of 9.1 meters, how fast must the roller coaster be moving at the

Verdich [7]

Answer:

at least 20mph

Explanation:

8 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago