You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large.

(a) What is the planet's radius?

Question

You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large.

(a) What is the planet's radius?

Accepted Answer

Everyone in this problem, we're told to suppose that there is a planet with a mass that is four times the mass of the earth. If a spaceship lands on the planet's surface, it measures the freefall acceleration to be 1/8 of that of the earth's acceleration due to gravity. And we're asked to determine the radius of this planet. We're given four answer choices. All of meters. Option A 20.3 times 10 to the exponent three. Option B 36 times 10 to the exponent three. Option C 20.3 times 10 to the exponent six and option D 36 times 10 to the exponent six. All right. So let's write down first what we were given. So we were told that the mass of this planet that we're dealing with is gonna be equal to four times the mass of the earth. OK. So we're gonna write MP, the mass of the planet is equal to four multiplied by MP, the mass of the earth. And similarly, sorry, the acceleration due to gravity on this planet. GP is gonna be equal to 1/8 the acceleration due to gravity on earth, which we'll call GE. So we're interested in finding the radius of this planet. OK. So we have the mass or we have some information about mass, we have some information about the acceleration due to gravity. We want to relate these to the radius. So recall that we have the equation G is equal to G multiplied by M divided by R squared. OK. So little G acceleration due to gravity is equal to G gravitational constant multiplied by the mass divided by R squared where R is the radius. So let's go ahead and write this out for our two situation. We have the earth and we have this other planet for the earth. We're gonna use great. If we write this for the earth, we can write that little GE the acceleration due to gravity on earth is equal to G gravitational constant multiplied by me, the mass of the earth divided by re squared, the radius of the earth. For this other planet, we can do the exact same thing. The subscripts are just gonna be peak. So we have GP is got to be multiplied by MP, divided by RP squared. All right. Now, we don't have the values for GP and MP, but we have a relationship to the earth. So we wanna find this radius of the planet RP. So we're gonna start by using this equation, we're gonna substitute what we know about the acceleration due to gravity in the mass on this planet. And see what we can do. So we know that the acceleration due to gravity is 1/8 of that occur. So we can substitute that in we have 1/8 G. Does he want to be you multiplied 54 me divided by re Oops RP squid. We know what Ge is and we've already written what GE is in terms of G and me. So let's substitute that value into our equation and see if we can get any of these terms to cancel is on the left hand side, we have 18 multiplied like capital G multiplied by me, divided by re square. And this is gonna be equal to capital multiplied me divided by RP squared. So we're just substituting what we know about the gravitational acceleration on earth. OK? And now you can notice that we can cancel out terms and we can divide both sides by G, we can divide both sides by me and both of those will divide out on each side. So one more left. OK? Is that 1/ are eastward? OK. With that re squared in the denominator is equal to four divided by RP squared. Now, recall we're looking for RP. So we're gonna multiply both sides by RP squared. We're gonna multiply both sides by eight re squared to get everything out of the denominators. And we can write that RP squared is gonna be equal to four multiplied by eight multiplied by re square. So now we have a relationship for the radius of this planet that is only in terms of the radius of earth. OK. So it didn't matter what the mass of earth was, it didn't matter what the gravitational constant was. All that matters is what the relationship between these radiuses is and what the relationship between those masses were. Now the radius of earth, this is a constant that you could look up in your textbook or in a table that your professor had provided. So we get that RP squared is equal to multiplied by 6.378 times to the exponent 6 m ward. And if we simplify, we get that the radius of this planet is about 36.0 times 10 to the exponent 6 m. So we get, we didn't need to know exactly the mass, exactly the acceleration due to gravity. We just needed to know the relationship between those on the planet we're interested in and those on earth going up and comparing our answer choices, we found that the answer was about 36 times 10 to the exponent 6 m which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helps see you in the next one.

You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of earth but the free-fall acceleration at the surface is only one-fourth as large. (a) What is the planet's radius?

Verified Solution

Key Concepts

Gravitational Force

Free-Fall Acceleration

Planetary Radius Calculation