A hypothetical planet has a radius 1.5 times that of Earth, but has the same mass.?

Question

What is the acceleration due to gravity near its surface?

PLEASE SHOW WORK. THANK YOU

Answer 1

The gravitational law states that the force of gravity is proportional to the mass, and inversely proportional to the square of the distance.

If the hypothetical planet has the same mass as the earth, then the force of gravity will be the same for a given distance to the center (this is a neat thing about those laws: it applies the same if all the mass of the planet was concentrated on a single point, and everything between the core and the surface was totally massless, provided that you are above the surface). Now, your distance to the center went from 1 Earth radius to 1.5. 1.5 square is 2.25, so the force and acceleration at the surface will be 1/2.25 the one of the Earth, or 0.4444 times what we have around here. Since Earth acceleration is 9.81 m/s^2, then that hypothetical planet would have an acceleration due to gravity at the surface of 4.36 m/s^2.

Answer 2

Gravitational acceleration is based only on mass not size.

A remember black holes have much less size than earth but have much more mass. Thus it has a larger gravitational pull.

But in this problem we are asked what is the gravitational acceleration near the surface on this fictitious planet. It would be the same acceleration you would feel if you were 1.5 times from the center of the earth.

F=GMm/r^2 Newton's law of Gravitiation

F = ma Newton's second law

ma = GMm/r^2 eliminate m and solve for a

a = GM/r^2. You know G, the universal gravitational constant, and M, the mass of the earth, place r = 1.5*(radius of the earth).

You are done. The aforementioned constants should be in the front or back cover of your book.

Answer 3

Let G be the gravitational constant 6.6742 × 10-11 N·m2/kg2
M1 be mass of Earth
M2 be mass of hypothetical planet
R1 be radius of Earth (average,neglect equatorial bulge, etc)
R2 be radius of hypothetical planet

"Gravitational acceleration" is the acceleration that an object experiences because of gravity when it falls freely close to the surface of a massive body, such as a planet. Also known as the acceleration of free fall, its value can be calculated from the formula

g = GM / [(R+h)(R+h)]

where M is the mass of the gravitating body (such as the Earth), R is the radius of the body, h is the height above the surface, and G is the gravitational constant. If the falling object is at, or very nearly at, the surface of the gravitating body, then the above equation reduces to

g = GM / (R)(R)

For the hypothetical planet, R2 = 1.5(R1) and M2=M1
Therefore the acceleration due to gravity near the surface =

g2 = (G(M2) / [3(R1)/2][3(R1)/2]
= = = (4/9)G(M2) / [(R1)(R1)]
= = = 0.4444(G)(M1)/[(R1)(R1)]

... from here, simply plug in the known values for the gravitational constant, earth's radius, and earth's mass to get a hard number.

Answer 4

easy

the acceleration due to gravity is just the gravitational force, divided by the mass of the lighter object

F = (G * M1 * M2) / R^2

G gravitational constant

M1 mass of planet

M2 mass of object

R radius of planet


So a = F / M2 = G * M1 / R^2

This is normally known as "g" and is worth 9.80 meters per second squared (32.2 feet per second squared)

Now your exoplanet has the same mass as Earth. But its radius is 1.5 times larger. So its radius, squared, is 1.5 * 1.5 = 2.25 times larger than Earth's.

So since all the rest is equal, the acceleration due to gravity near the surface will be 2.25 times less than g. In numbers, this is 4.36 meters per second squared (or 14.3 feet per second squared).


Good luck

Answer 5

F = G*M*m/r2
= 6.67*10^-11 m3/s2kg * 6*10^24 kg * m / 1.5*4*10^12 m2

or, more simply 9.8/1.5^2
=4.36