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Aleksandr-060686 [28]

3 years ago

11

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

ds? (b) What is the period in Earth years? Assume that one Earth year is exactly 365 days, with 24 hours in each day.

2 answers:

dedylja [7]3 years ago

7 0

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

or

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

Tanya [424]3 years ago

6 0

Answer:

(a) 3.74 x 10^8 seconds

(b) 11.86 earth year

Explanation:

Radius, R = 7.78 x 10^11 m

mass of sun, M = 1.99 x 10^30 kg

(a) Let T be the period of Jupiter.

$T=2\pi \sqrt{\frac{R^{3}}{GM}}$

$T=2\pi \sqrt{\frac{\left ( 7.78\times10^{11} \right )^{3}}{6.67\times10^{-11}\times1.99\times10^{30}}}$

T = 3.74 x 10^8 seconds

(b) 1 year = 365 days

1 day = 24 hour

1 hour = 60 minutes

1 minute = 60 seconds

T = 11.86 earth year.

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Answer:

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Explanation:

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$0\ = \ 2.45 \ m \ + \ 0 \ * \ t \ - \frac{1}{2} \ g \ t_{fall}^2$

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$\ 2.45 \ m \ = \frac{1}{2} \ g \ t_{fall}^2$

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$\frac{ 2 \ * \ 2.45 \ m}{g} \ = t_{fall}^2$

$\sqrt{ \frac{ 2 \ * \ 2.45 \ m}{g} } \ = \ t_{fall}$

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$t_{fall} \ = \sqrt{ \frac{ 2 \ * \ 2.45 \ m}{9.8 \frac{m}{s^2} } }$

$t_{fall} \ = \sqrt{ \frac{1}{2} }$

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$d_h \ = \ 7.78 \ m$

And this must be the distance we are looking for. So, the saddle and the limb must be at 7.78 meters when the man makes his move.

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