SECTION A. Planet discovery Lets imagine a planet at distance of 3 Au (1 Au 1.50

Question

SECTION A. Planet discovery Lets imagine a planet at distance of 3 Au (1 Au 1.50 x heavier than the earth (ME 5.98 x 1024 kg) and its radius is twice larger than the ea = 6.37x 104 m). Universal gravitational constant G-667 x 10-11 m3 kg-'s- 1011 m) from the Sun. It is twice A-1. What is the period of this new planet orbiting the Sun? Answer: ears A-2. What is the gravitational acceleration (g) on this planet? (Earth's gravitational acceleration is 9.8 m/s) Answer: m/sz A-3. If you throw a 10-kg object at 6000 m/s initial speed from the ground, how high (altitude) does the object go on this planet? Answer:

Explanation / Answer

part A:

from Kepler law,

T^2 = 4pi^2 R^3/GM

T^2 = 4*3.14*3.14 * (2* 6.37*10^6)^3/(6.67*10^-11 * 2* 5.98 *10^24)

T = 10110 secs or 2.8 hrs

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part B:

Accleration Due to gravity a = GM/r^2

G is universal gravitatioanl constant = 6.67 *10^-11 Nm^2/kg^2

M is mass of the planet

here M = 2* ME

radius r = 2Re

so

g = G*(2Me)/(2Re)^2

g = 2 GMe/4Re^2

g = 1/2 * GM/Re^2

GM/Re^2 is g on earth = 9.81m/s^2

so

g on planet = 1/2* 9.81 = 4.905 m/s^2
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part C:

use the formula from Energy considerations

0.5 mv^2   = mgh

v^2 = 2 g h

h = V^2/2g

h = 6000^2/(2 * 4.905)

h = 3.67 *10^6 m    or 3670 km

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A4:

from KEplers third law

T^2 = 4pi^2 R^3/GM

R^3 = GMT^2/4pi^2

R^3 = (6.67*10^-11 * 2* 5.98 *10^24 * 43200^2)/(4*3.14^2)

R = 3.35 *10^7 m from the center of the planet :

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part 5:

Vesc = sqrt2* gR

Vesc^2 = 1.414 * 4.905 * 2* 6.37 *10^6

Vesc = 9.4 km/s

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